Exercise 0.1: Computing in algebraic form: ten direct calculations

For each line, carry out the calculation and write the result in the algebraic form $a+bi$ , then give what is asked: most often the real part $a$ and the imaginary part $b$ , which is a real number.

Compute as with polynomials in $i$ , then replace $i^{2}$ by $-1$ and collect the real parts and the imaginary parts. To divide, multiply the numerator and the denominator by the conjugate of the denominator, which becomes real: $(a+bi)(a-bi)=a^{2}+b^{2}$ . The modulus is $|a+bi|=\sqrt{a^{2}+b^{2}}$ , and a power of $i$ reduces to the remainder of the exponent on division by $4$ .

a) $(5-3i)+(-1+7i)$ , then $(5-3i)-(-1+7i)$ : give the imaginary part of each.
b) $(2-5i)(3+4i)$ .
c) $(4-3i)^{2}$ .
d) $i^{58}+i^{103}$ .
e) $|-8+15i|$ and $|6-2i|$ .
f) $\overline{(2+3i)(1-2i)}$ .
g) $\frac{1}{1-3i}$ .
h) $\frac{7+i}{1-2i}$ .
i) $(1+2i)^{2}\,\overline{(1-i)}+\frac{5i}{1+2i}$ .
j) The solutions of $2z^{2}-2z+5=0$ in $\mathbb{C}$ : give the real part and the imaginary part of the one whose imaginary part is positive.

Type your answers, the page tells you right or wrong 0/20

a)

Imaginary part of the sum

Imaginary part of the difference

b)

\operatorname{Re}

\operatorname{Im}

c)

\operatorname{Re}

\operatorname{Im}

d)

\operatorname{Re}

\operatorname{Im}

e)

|-8+15i|

|6-2i|

f)

\operatorname{Re}

\operatorname{Im}

g)

\operatorname{Re}

\operatorname{Im}

h)

\operatorname{Re}

\operatorname{Im}

i)

\operatorname{Re}

\operatorname{Im}

j)

\operatorname{Re}(z)

\operatorname{Im}(z)

Show the solution

Answers

a) $4+4i$ and $6-10i$ ; $4$ and $-10$
b) $26-7i$
c) $7-24i$
d) $-1-i$
e) $17$ ; $2\sqrt{10}\approx 6.325$
f) $\overline{8-i}=8+i$
g) $\frac{1}{10}+\frac{3}{10}i$
h) $1+3i$
i) $-5+2i$
j) $z=\frac{1}{2}\pm\frac{3}{2}i$ ; $\frac{1}{2}$ and $\frac{3}{2}$

The move that decides everything: replace $i^{2}$ by $-1$ as soon as it appears, then collect the real part and the imaginary part. For a division, you never divide by a complex number: multiply the top and the bottom by the conjugate of the denominator, which becomes real.

a)

$(5-3i)+(-1+7i)=4+4i$ and $(5-3i)-(-1+7i)=5-3i+1-7i=6-10i$ : imaginary parts $4$ and $-10$ . Pitfall: distributing the minus sign over the $-1$ only, which gives $6+4i$ .

b)

$(2-5i)(3+4i)=6+8i-15i-20i^{2}=6+20-7i=26-7i$ . Pitfall: reading $-20i^{2}$ as $-20$ , hence $-14-7i$ .

c)

$(4-3i)^{2}=16-24i+9i^{2}=16-9-24i=7-24i$ . Pitfall: $16+9=25$ , forgetting that $i^{2}=-1$ .

d)

Divide the exponent by $4$ : $58=4\times 14+2$ , so $i^{58}=i^{2}=-1$ ; $103=4\times 25+3$ , so $i^{103}=i^{3}=-i$ . The sum is $-1-i$ .

e)

$|-8+15i|=\sqrt{64+225}=\sqrt{289}=17$ and $|6-2i|=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}\approx 6.325$ . Pitfall: $\sqrt{36-4}$ , squaring $-2i$ instead of $-2$ ; the modulus uses real numbers only.

f)

The product first: $(2+3i)(1-2i)=2-4i+3i-6i^{2}=8-i$ , whose conjugate is $8+i$ . The same result comes from $\overline{2+3i}\cdot\overline{1-2i}=(2-3i)(1+2i)=2+4i-3i-6i^{2}=8+i$ : the conjugate of a product is the product of the conjugates.

g)

$\frac{1}{1-3i}=\frac{1+3i}{(1-3i)(1+3i)}=\frac{1+3i}{1+9}=\frac{1}{10}+\frac{3}{10}i$ . Pitfall: writing $(1-3i)(1+3i)=1-9$ , when in fact $-(3i)^{2}=+9$ .

h)

$\frac{7+i}{1-2i}=\frac{(7+i)(1+2i)}{(1-2i)(1+2i)}=\frac{7+14i+i+2i^{2}}{1+4}=\frac{5+15i}{5}=1+3i$ .

i)

$(1+2i)^{2}=1+4i+4i^{2}=-3+4i$ and $\overline{1-i}=1+i$ , so $(-3+4i)(1+i)=-3-3i+4i+4i^{2}=-7+i$ . Then $\frac{5i}{1+2i}=\frac{5i(1-2i)}{1+4}=i-2i^{2}=2+i$ . Total: $-7+i+2+i=-5+2i$ .

j)

$\Delta=(-2)^{2}-4\times 2\times 5=4-40=-36<0$ : two complex conjugate solutions, with $\sqrt{\Delta}=6i$ . So $z=\frac{2\pm 6i}{4}=\frac{1}{2}\pm\frac{3}{2}i$ ; the one with positive imaginary part is $\frac{1}{2}+\frac{3}{2}i$ . Pitfall: dividing by $2$ instead of $2a=4$ .

Check: multiply back. In h), $(1+3i)(1-2i)$ must give $7+i$ again; in g), $(1-3i)\left(\frac{1}{10}+\frac{3}{10}i\right)=1$ ; in j), the sum of the two solutions is $1=-\frac{b}{a}$ and their product is $\frac{1}{4}+\frac{9}{4}=\frac{5}{2}=\frac{c}{a}$ .

Tick the exercises you have done or want to review: a free account, no password, keeps your ticks and your right answers from one visit to the next, tells you which chapter to tackle next and lets you ask for the exercise you are missing. Create your space, an email is enough.

Exercise 0.2: Converting to exponential form and computing powers: nine direct calculations

For each line, compute what is asked. Arguments are principal arguments, in $(-\pi, \pi]$ , and are entered as multiples of $\pi$ : the argument $-\frac{3\pi}{4}$ is typed $-3/4$ .

The modulus first, $r=\sqrt{a^{2}+b^{2}}$ , then the argument from BOTH conditions $\cos\theta=\frac{a}{r}$ and $\sin\theta=\frac{b}{r}$ , after locating the point in its quadrant. In exponential form, the moduli multiply and the arguments add; De Moivre's theorem gives $\left(re^{i\theta}\right)^{n}=r^{n}e^{in\theta}$ , for negative $n$ as well. Bring the final argument back into $(-\pi, \pi]$ before returning to algebraic form.

a) $z=-7$ and $w=-3i$ : give $\arg(z)$ and $\arg(w)$ .
b) $z=3\sqrt{3}+3i$ : give $|z|$ and $\arg(z)$ .
c) $z=-\sqrt{2}-\sqrt{2}\,i$ : give $|z|$ and $\arg(z)$ .
d) $z=6e^{i\frac{5\pi}{6}}$ : write $z$ in algebraic form.
e) $2e^{i\frac{\pi}{12}}\times 5e^{i\frac{\pi}{4}}$ : write the product in exponential form, then in algebraic form; give its modulus and its imaginary part.
f) $\frac{4e^{i\frac{2\pi}{3}}}{2e^{-i\frac{\pi}{2}}}$ : write the quotient in exponential form; give its modulus and its principal argument.
g) $(\sqrt{3}-i)^{8}$ , by De Moivre's theorem, in algebraic form.
h) $(1-i)^{-6}$ : give its principal argument and its imaginary part.
i) The cube root of $-27$ obtained for $k=0$ in the formula $w_{k}=\sqrt[3]{r}\,e^{i\frac{\theta+2k\pi}{3}}$ , with $\theta\in(-\pi, \pi]$ : give its real part and its imaginary part.

Type your answers, the page tells you right or wrong 0/18

a)

\arg(-7)

(in multiples of π)

\arg(-3i)

(in multiples of π)

b)

|z|

\arg(z)

(in multiples of π)

c)

|z|

\arg(z)

(in multiples of π)

d)

\operatorname{Re}(z)

\operatorname{Im}(z)

e)

Modulus

\operatorname{Im}

f)

Modulus

Principal argument (in multiples of π)

g)

\operatorname{Re}

\operatorname{Im}

h)

Principal argument (in multiples of π)

\operatorname{Im}

i)

\operatorname{Re}(w_{0})

\operatorname{Im}(w_{0})

Show the solution

Answers

a) $\arg(-7)=\pi$ ; $\arg(-3i)=-\frac{\pi}{2}$
b) $z=6e^{i\frac{\pi}{6}}$ ; $6$ and $\frac{\pi}{6}$
c) $z=2e^{-i\frac{3\pi}{4}}$ ; $2$ and $-\frac{3\pi}{4}$
d) $z=-3\sqrt{3}+3i\approx-5.196+3i$
e) $10e^{i\frac{\pi}{3}}=5+5\sqrt{3}\,i$ ; $10$ and $5\sqrt{3}\approx 8.660$
f) $2e^{-i\frac{5\pi}{6}}$ ; $2$ and $-\frac{5\pi}{6}$
g) $256e^{i\frac{2\pi}{3}}=-128+128\sqrt{3}\,i$ ; $-128$ and $\approx 221.703$
h) $\frac{1}{8}e^{-i\frac{\pi}{2}}=-\frac{1}{8}i$ ; $-\frac{\pi}{2}$ and $-0.125$
i) $w_{0}=3e^{i\frac{\pi}{3}}=\frac{3}{2}+\frac{3\sqrt{3}}{2}i$ ; $1.5$ and $\approx 2.598$

The move that decides everything: place the point in the plane BEFORE looking for the argument. The quadrant fixes the sign and the value of the angle, which the arctangent alone cannot tell apart; after that, in exponential form, everything comes down to multiplying moduli and adding arguments.

a)

$-7$ lies on the negative real half-axis: $|-7|=7$ and $\arg(-7)=\pi$ . $-3i$ lies on the negative imaginary half-axis: $|-3i|=3$ and $\arg(-3i)=-\frac{\pi}{2}$ . Pitfall: giving $\arg(-7)=0$ because the number is real; a negative real points to the left.

b)

$|z|=\sqrt{27+9}=\sqrt{36}=6$ , $\cos\theta=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}$ and $\sin\theta=\frac{3}{6}=\frac{1}{2}$ : first quadrant, $\theta=\frac{\pi}{6}$ . So $z=6e^{i\frac{\pi}{6}}$ .

c)

$|z|=\sqrt{2+2}=2$ , $\cos\theta=-\frac{\sqrt{2}}{2}$ and $\sin\theta=-\frac{\sqrt{2}}{2}$ : third quadrant, $\theta=-\frac{3\pi}{4}$ , and $z=2e^{-i\frac{3\pi}{4}}$ . Pitfall: $\arctan\frac{-\sqrt{2}}{-\sqrt{2}}=\arctan 1=\frac{\pi}{4}$ , which is the argument of the OPPOSITE point.

d)

$6e^{i\frac{5\pi}{6}}=6\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)=6\left(-\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=-3\sqrt{3}+3i$ , that is a real part $-3\sqrt{3}\approx-5.196$ and an imaginary part $3$ . Pitfall: $\cos\frac{5\pi}{6}$ is negative, the angle is in the second quadrant.

e)

The moduli multiply, the arguments add: $2\times 5=10$ and $\frac{\pi}{12}+\frac{\pi}{4}=\frac{\pi}{12}+\frac{3\pi}{12}=\frac{\pi}{3}$ . So the product is $10e^{i\frac{\pi}{3}}=5+5\sqrt{3}\,i$ , with modulus $10$ and imaginary part $5\sqrt{3}\approx 8.660$ .

f)

The moduli divide, the arguments subtract: $\frac{4}{2}=2$ and $\frac{2\pi}{3}-\left(-\frac{\pi}{2}\right)=\frac{7\pi}{6}$ . This angle exceeds $\pi$ : subtract $2\pi$ , which gives the principal argument $\frac{7\pi}{6}-2\pi=-\frac{5\pi}{6}$ , and the quotient is $2e^{-i\frac{5\pi}{6}}$ . Pitfall: subtracting $\frac{\pi}{2}$ instead of $-\frac{\pi}{2}$ , which gives $\frac{\pi}{6}$ .

g)

$\sqrt{3}-i=2e^{-i\frac{\pi}{6}}$ , in the fourth quadrant. By De Moivre's theorem, $(\sqrt{3}-i)^{8}=2^{8}e^{-i\frac{8\pi}{6}}=256e^{-i\frac{4\pi}{3}}=256e^{i\frac{2\pi}{3}}$ , adding $2\pi$ . So $(\sqrt{3}-i)^{8}=256\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=-128+128\sqrt{3}\,i$ , with $128\sqrt{3}\approx 221.703$ . Pitfall: forgetting to raise the modulus to the power $8$ , which gives $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ .

h)

$1-i=\sqrt{2}\,e^{-i\frac{\pi}{4}}$ , so $(1-i)^{-6}=(\sqrt{2})^{-6}e^{i\frac{6\pi}{4}}=\frac{1}{8}e^{i\frac{3\pi}{2}}=\frac{1}{8}e^{-i\frac{\pi}{2}}=-\frac{1}{8}i$ : principal argument $-\frac{\pi}{2}$ , imaginary part $-\frac{1}{8}=-0.125$ . Pitfall: the negative exponent ALSO changes the sign of the argument, $-6\times\left(-\frac{\pi}{4}\right)=+\frac{3\pi}{2}$ . Cross-check: $(1-i)^{2}=-2i$ , so $(1-i)^{6}=(-2i)^{3}=8i$ and $\frac{1}{8i}=-\frac{i}{8}$ .

i)

$-27=27e^{i\pi}$ , so $w_{k}=3e^{i\frac{\pi+2k\pi}{3}}$ . For $k=0$ : $w_{0}=3e^{i\frac{\pi}{3}}=\frac{3}{2}+\frac{3\sqrt{3}}{2}i$ , that is a real part $1.5$ and an imaginary part $\frac{3\sqrt{3}}{2}\approx 2.598$ . Pitfall: answering $-3$ , the real root, which is $w_{1}$ ; the root for $k=0$ is real only when the starting argument is zero.

Check: an exponential form is cross-checked by returning to algebraic form, $re^{i\theta}=r\cos\theta+ir\sin\theta$ , and a power or a root is checked by multiplying back: here $w_{0}^{3}=27e^{i\pi}=-27$ , and in g) the modulus $|-128+128\sqrt{3}\,i|=128\times 2=256=2^{8}$ .

Exercise 1: Algebraic form, conjugate and division

Consider the complex numbers $z_{1}=3+2i$ and $z_{2}=1-4i$ .

a) Compute $z_{1}+z_{2}$ and $z_{1}z_{2}$ in algebraic form.
b) Give $\bar{z_{1}}$ and $\bar{z_{2}}$ , then compute $|z_{1}|$ and $|z_{2}|$ .
c) Write $\dfrac{z_{1}}{z_{2}}$ in algebraic form.
d) Compute $z_{1}\bar{z_{1}}$ and comment on the result. Then find $i^{2027}$ .

Show the solution

Answers

a) $4-2i$ ; $11-10i$
b) $3-2i$ , $1+4i$ ; $\sqrt{13}$ , $\sqrt{17}$
c) $-\frac{5}{17}+\frac{14}{17}i$
d) $z_1\bar{z_1}=13$ ; $i^{2027}=-i$

a)

$z_{1}+z_{2}=(3+1)+(2-4)i=4-2i$ . For the product, expand and then use $i^{2}=-1$ : $z_{1}z_{2}=(3+2i)(1-4i)=3-12i+2i-8i^{2}=3-10i+8=11-10i$ . The term $-8i^{2}$ becomes $+8$ , and that sign change is exactly what gets forgotten under pressure.

b)

$\bar{z_{1}}=3-2i$ and $\bar{z_{2}}=1+4i$ . The moduli: $|z_{1}|=\sqrt{3^{2}+2^{2}}=\sqrt{13}$ and $|z_{2}|=\sqrt{1^{2}+(-4)^{2}}=\sqrt{17}$ . The modulus uses the squares of the components, so the sign of the $-4$ disappears.

c)

Multiply the top and the bottom by the conjugate of the denominator: $\dfrac{z_{1}}{z_{2}}=\dfrac{(3+2i)(1+4i)}{(1-4i)(1+4i)}$ . Numerator: $3+12i+2i+8i^{2}=3+14i-8=-5+14i$ . Denominator: $1^{2}+4^{2}=17$ , a real number, which is the whole point of the operation. So $\dfrac{z_{1}}{z_{2}}=\dfrac{-5+14i}{17}=-\dfrac{5}{17}+\dfrac{14}{17}i$ .

Check with the moduli: $\left|\dfrac{z_{1}}{z_{2}}\right|=\dfrac{\sqrt{13}}{\sqrt{17}}\approx 0.874$ , and directly $\dfrac{\sqrt{25+196}}{17}=\dfrac{\sqrt{221}}{17}\approx 0.874$ . Consistent.

d)

$z_{1}\bar{z_{1}}=(3+2i)(3-2i)=9-6i+6i-4i^{2}=9+4=13$ . The result is a REAL number, equal to $|z_{1}|^{2}=(\sqrt{13})^{2}=13$ . This is the property $z\bar{z}=|z|^{2}$ , and it is what makes the division in part c) work.

For $i^{2027}$ : the powers of $i$ repeat with period $4$ , since $i^{1}=i$ , $i^{2}=-1$ , $i^{3}=-i$ , $i^{4}=1$ , and then the cycle starts again. So divide the exponent by $4$ : $2027=4\times 506+3$ , the remainder is $3$ . Therefore $i^{2027}=i^{3}=-i$ . No calculator is needed, only the remainder on division by $4$ matters.

Exercise 2: Modulus, argument and exponential form

Consider the complex number $z=-2+2i\sqrt{3}$ .

a) Compute $|z|$ , then find $\arg(z)$ in radians.
b) Write $z$ in trigonometric form and in exponential form.
c) Give $\bar{z}$ and $-z$ in exponential form, without going back to the algebraic form.
d) Let $w=1+i$ . Write $w$ in exponential form, then compute $zw$ in two ways, in exponential form and in algebraic form, and check that the moduli agree.

Show the solution

Answers

a) $|z|=4$ , $\arg z=\frac{2\pi}{3}$
b) $4e^{i\frac{2\pi}{3}}$
c) $\bar z=4e^{-i\frac{2\pi}{3}}$ , $-z=4e^{-i\frac\pi3}$
d) $zw=4\sqrt2\,e^{i\frac{11\pi}{12}}$

a)

$|z|=\sqrt{(-2)^{2}+(2\sqrt{3})^{2}}=\sqrt{4+12}=\sqrt{16}=4$ . For the argument, impose BOTH conditions: $\cos\theta=\dfrac{-2}{4}=-\dfrac{1}{2}$ and $\sin\theta=\dfrac{2\sqrt{3}}{4}=\dfrac{\sqrt{3}}{2}$ . A negative cosine and a positive sine put the angle in the second quadrant, and the reference angle for $\dfrac{1}{2}$ and $\dfrac{\sqrt{3}}{2}$ is $\dfrac{\pi}{3}$ . So $\theta=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}$ .

The classic pitfall: using the calculator for $\arctan\left(\dfrac{2\sqrt{3}}{-2}\right)=\arctan(-\sqrt{3})=-\dfrac{\pi}{3}$ , and stopping there. That value is in the fourth quadrant, while $z$ is in the second. The arctangent function only ever returns angles between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ : it cannot tell opposite quadrants apart. Always locate the point before concluding.

b)

Trigonometric form: $z=4\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$ . Exponential form: $z=4e^{i\frac{2\pi}{3}}$ .

c)

The conjugate has the same modulus and the opposite argument: $\bar{z}=4e^{-i\frac{2\pi}{3}}$ . Geometrically, it is the reflection in the real axis.

For $-z$ , multiplying by $-1$ means multiplying by $e^{i\pi}$ , that is adding $\pi$ to the argument: $-z=4e^{i\left(\frac{2\pi}{3}+\pi\right)}=4e^{i\frac{5\pi}{3}}$ , which is more conveniently written $4e^{-i\frac{\pi}{3}}$ after subtracting $2\pi$ . Geometrically, it is the point reflection through the origin, a rotation by a half-turn.

d)

$|w|=\sqrt{1+1}=\sqrt{2}$ , and the point $(1, 1)$ lies on the bisector of the first quadrant, so $\arg(w)=\dfrac{\pi}{4}$ and $w=\sqrt{2}\,e^{i\frac{\pi}{4}}$ .

In exponential form: $zw=4\sqrt{2}\,e^{i\left(\frac{2\pi}{3}+\frac{\pi}{4}\right)}$ . With a common denominator: $\dfrac{2\pi}{3}+\dfrac{\pi}{4}=\dfrac{8\pi}{12}+\dfrac{3\pi}{12}=\dfrac{11\pi}{12}$ . So $zw=4\sqrt{2}\,e^{i\frac{11\pi}{12}}$ , in three lines.

In algebraic form: $zw=(-2+2i\sqrt{3})(1+i)=-2-2i+2i\sqrt{3}+2i^{2}\sqrt{3}=-2-2i+2\sqrt{3}\,i-2\sqrt{3}=(-2-2\sqrt{3})+(2\sqrt{3}-2)i$ .

Checking the moduli: $|zw|=\sqrt{(2+2\sqrt{3})^{2}+(2\sqrt{3}-2)^{2}}=\sqrt{(4+8\sqrt{3}+12)+(12-8\sqrt{3}+4)}=\sqrt{32}=4\sqrt{2}$ . The terms in $\sqrt{3}$ cancel exactly, and we do get $|z||w|=4\times\sqrt{2}$ . Both forms describe the same number, but the exponential form gave the result without a single expansion.

Exercise 3: De Moivre's theorem and powers

Consider the complex number $z=1+i\sqrt{3}$ .

a) Write $z$ in exponential form.
b) Compute $z^{6}$ using De Moivre's theorem.
c) Compute $z^{10}$ and give the result in algebraic form.
d) Compute $z^{-3}$ in algebraic form, then check that $z^{3}\times z^{-3}=1$ .

Show the solution

Answers

a) $z=2e^{i\frac\pi3}$
b) $z^6=64$
c) $z^{10}=-512-512\sqrt3\,i$
d) $z^{-3}=-\frac18$

a)

$|z|=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=2$ . Then $\cos\theta=\dfrac{1}{2}$ and $\sin\theta=\dfrac{\sqrt{3}}{2}$ , both positive, so first quadrant: $\theta=\dfrac{\pi}{3}$ . Thus $z=2e^{i\frac{\pi}{3}}$ .

b)

By De Moivre's theorem: $z^{6}=2^{6}e^{i\cdot 6\cdot\frac{\pi}{3}}=64\,e^{i2\pi}$ . But $e^{i2\pi}=\cos 2\pi+i\sin 2\pi=1+0i=1$ . So $z^{6}=64$ , a positive real number. Compare with the work that expanding $(1+i\sqrt{3})^{6}$ with the binomial theorem would have taken: seven terms, with powers of $\sqrt{3}$ and of $i$ to keep track of. The exponential form settles the question in one line.

c)

$z^{10}=2^{10}e^{i\cdot 10\cdot\frac{\pi}{3}}=1024\,e^{i\frac{10\pi}{3}}$ . Bring the argument back into $(-\pi, \pi]$ by subtracting $4\pi=\dfrac{12\pi}{3}$ : $\dfrac{10\pi}{3}-\dfrac{12\pi}{3}=-\dfrac{2\pi}{3}$ . So $z^{10}=1024\,e^{-i\frac{2\pi}{3}}$ .

In algebraic form: $\cos\left(-\dfrac{2\pi}{3}\right)=-\dfrac{1}{2}$ and $\sin\left(-\dfrac{2\pi}{3}\right)=-\dfrac{\sqrt{3}}{2}$ (third quadrant, both negative). So $z^{10}=1024\left(-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\right)=-512-512\sqrt{3}\,i$ .

d)

De Moivre's theorem also holds for negative exponents: $z^{-3}=2^{-3}e^{-i\pi}=\frac{1}{8}\times(-1)=-\frac{1}{8}$ . Check: $z^{3}=8e^{i\pi}=-8$ , and $(-8)\times\left(-\frac{1}{8}\right)=1$ . Computing $\frac{1}{(1+i\sqrt{3})^{3}}$ directly would have taken an expansion and a division by the conjugate.

Exercise 4: The nth roots of a complex number

We want to solve the equation $z^{3}=8i$ in $\mathbb{C}$ .

a) Write $8i$ in exponential form.
b) Find the three solutions of the equation in exponential form, then in algebraic form.
c) Check by direct calculation that $\sqrt{3}+i$ is indeed a solution.
d) Plot the three solutions in the complex plane and describe the figure they form. What is their sum?

Show the solution

Answers

a) $8i=8e^{i\frac\pi2}$
b) $\sqrt3+i$ , $-\sqrt3+i$ , $-2i$
c) $(\sqrt3+i)^3=8i$
d) Equilateral triangle, zero sum

a)

The number $8i$ has real part $0$ and imaginary part $8$ , so it lies on the positive imaginary half-axis. Its modulus is $8$ and its argument $\dfrac{\pi}{2}$ . So $8i=8e^{i\frac{\pi}{2}}$ .

b)

The cube roots have modulus $8^{1/3}=2$ and arguments $\dfrac{\frac{\pi}{2}+2k\pi}{3}=\dfrac{\pi}{6}+\dfrac{2k\pi}{3}$ for $k=0, 1, 2$ .

For $k=0$ : $z_{0}=2e^{i\frac{\pi}{6}}=2\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)=\sqrt{3}+i$ .

For $k=1$ : the argument is $\dfrac{\pi}{6}+\dfrac{2\pi}{3}=\dfrac{\pi}{6}+\dfrac{4\pi}{6}=\dfrac{5\pi}{6}$ , so $z_{1}=2e^{i\frac{5\pi}{6}}=2\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)=-\sqrt{3}+i$ .

For $k=2$ : the argument is $\dfrac{\pi}{6}+\dfrac{4\pi}{3}=\dfrac{\pi}{6}+\dfrac{8\pi}{6}=\dfrac{9\pi}{6}=\dfrac{3\pi}{2}$ , so $z_{2}=2e^{i\frac{3\pi}{2}}=2(0-i)=-2i$ .

The mistake to avoid: stopping at the first root. An equation of degree $3$ has exactly three solutions in $\mathbb{C}$ , and the parameter $k$ must run through $0$ , $1$ and $2$ . Taking $k=3$ would give $z_{0}$ again, since it adds $2\pi$ to the argument.

c)

$(\sqrt{3}+i)^{2}=3+2\sqrt{3}\,i+i^{2}=3+2\sqrt{3}\,i-1=2+2\sqrt{3}\,i$ . Then $(\sqrt{3}+i)^{3}=(2+2\sqrt{3}\,i)(\sqrt{3}+i)=2\sqrt{3}+2i+2\cdot 3\,i+2\sqrt{3}\,i^{2}=2\sqrt{3}+2i+6i-2\sqrt{3}=8i$ . The check is conclusive.

d)

The three roots all have the same modulus $2$ , so they lie on the circle of radius $2$ centred at the origin. Their arguments are $\dfrac{\pi}{6}$ , $\dfrac{5\pi}{6}$ and $\dfrac{3\pi}{2}$ , evenly spaced by $\dfrac{2\pi}{3}$ , that is $120^{\circ}$ . They are therefore the vertices of an EQUILATERAL TRIANGLE inscribed in that circle.

Their sum: $(\sqrt{3}+i)+(-\sqrt{3}+i)+(-2i)=(\sqrt{3}-\sqrt{3})+(1+1-2)i=0$ . The sum is zero. This is no coincidence: the $n$ $n$ th roots of a complex number are the vertices of a regular polygon centred at the origin, and the centroid of a regular polygon is its centre, so their sum is always zero as soon as $n\geq 2$ . Another, algebraic reading: these roots are the solutions of $z^{3}-8i=0$ , a polynomial whose coefficient of $z^{2}$ is zero, and by the relations between coefficients and roots, the sum of the roots equals that coefficient up to sign, so $0$ . This zero sum is an excellent way to check your roots at the end of a calculation.

Exercise 5: Square roots of a complex number by the algebraic method

We look for the square roots of a complex number without going through the exponential form, by solving $(x+iy)^{2}=z$ directly. The method combines three real equations.

a) Find the square roots of $z=3+4i$ : set $(x+iy)^{2}=3+4i$ , write the system in $x$ and $y$ , adding the modulus equation $x^{2}+y^{2}=|z|$ , and solve it.
b) Find the square roots of $z=-5+12i$ in the same way.
c) Check each of your results by squaring, and explain why a non-zero complex number always has exactly two square roots, opposite to each other.
d) Application: solve the equation $w^{2}=3+4i$ in $\mathbb{C}$ , then explain how this method solves any quadratic equation with complex coefficients.

Show the solution

Answers

a) $\pm(2+i)$
b) $\pm(2+3i)$
c) Two opposite roots
d) $w=\pm(2+i)$ ; complex discriminant

a)

Set $(x+iy)^{2}=x^{2}-y^{2}+2xyi=3+4i$ , which gives $\begin{cases} x^{2}-y^{2}=3 \\ 2xy=4 \end{cases}$ . Add the modulus equation: $|(x+iy)^{2}|=|3+4i|$ gives $x^{2}+y^{2}=\sqrt{3^{2}+4^{2}}=5$ . Adding this to $x^{2}-y^{2}=3$ : $2x^{2}=8$ , so $x^{2}=4$ and $x=\pm 2$ ; subtracting: $2y^{2}=2$ , so $y^{2}=1$ and $y=\pm 1$ . The condition $2xy=4>0$ forces $x$ and $y$ to have the SAME sign. The square roots of $3+4i$ are therefore $2+i$ and $-2-i$ .

b)

Set $(x+iy)^{2}=-5+12i$ , so $\begin{cases} x^{2}-y^{2}=-5 \\ 2xy=12 \end{cases}$ and $x^{2}+y^{2}=|-5+12i|=\sqrt{25+144}=13$ . Adding this to $x^{2}-y^{2}=-5$ : $2x^{2}=8$ , so $x=\pm 2$ ; subtracting: $2y^{2}=18$ , so $y=\pm 3$ . The condition $2xy=12>0$ forces the same sign. The square roots of $-5+12i$ are $2+3i$ and $-2-3i$ .

c)

Check: $(2+i)^{2}=4+4i+i^{2}=4+4i-1=3+4i$ ✓, and $(2+3i)^{2}=4+12i+9i^{2}=4+12i-9=-5+12i$ ✓. The two roots of the same number are always opposite, because if $r^{2}=z$ then $(-r)^{2}=r^{2}=z$ as well, and a quadratic equation has no more than two solutions. A non-zero complex number therefore has exactly two square roots, symmetric about the origin. This is the difference with positive reals, where by convention only the positive root is kept; in $\mathbb{C}$ , neither of the two is preferred.

d)

The solutions of $w^{2}=3+4i$ are exactly the square roots found in a): $w=2+i$ or $w=-2-i$ . The method is general: to solve $aw^{2}+bw+c=0$ with complex coefficients, compute the discriminant $\Delta=b^{2}-4ac$ , which is a complex number, find a square root $\delta$ of it by the method above, and the solutions are $w=\dfrac{-b\pm\delta}{2a}$ . The only new step compared with the real case is extracting a square root of a complex number, and that is exactly what this exercise teaches.

Exercise 6: Polynomial equations and conjugate roots

This exercise is about solving polynomial equations in $\mathbb{C}$ and the property that governs their non-real roots.

a) Solve the equation $z^{2}-4z+13=0$ in $\mathbb{C}$ .
b) Prove that if a polynomial with REAL coefficients has $z_{0}$ as a root, then $\bar{z_{0}}$ is also a root.
c) Let $P(z)=z^{3}-5z^{2}+17z-13$ . Check that $1$ is a root of $P$ , then factor $P$ and find all its roots in $\mathbb{C}$ .
d) Prove that $|zw|=|z|\,|w|$ for all complex numbers $z$ and $w$ , then that $\overline{z+w}=\bar{z}+\bar{w}$ . Finally, explain with a counterexample why the result of b) would fail if the coefficients were not real.

Show the solution

Answers

a) $z=2\pm 3i$
b) Real coefficients: $P(\bar{z_0})=\overline{P(z_0)}=0$
c) $1$ , $2+3i$ , $2-3i$
d) $Q(z)=z-i$ : $-i$ is not a root

a)

The discriminant is $\Delta=(-4)^{2}-4(1)(13)=16-52=-36$ . It is negative, so there is no real solution, but there are two complex conjugate solutions. Write $\sqrt{-36}=6i$ , so $z=\dfrac{4\pm 6i}{2}=2\pm 3i$ . The solutions are $z=2+3i$ and $z=2-3i$ .

Check on the first one: $(2+3i)^{2}=4+12i+9i^{2}=4+12i-9=-5+12i$ , then $-5+12i-4(2+3i)+13=-5+12i-8-12i+13=0$ . It works.

b)

Let $P(z)=a_{n}z^{n}+\dots+a_{1}z+a_{0}$ with all the $a_{k}$ real, and suppose $P(z_{0})=0$ . Take the conjugate of both sides: $\overline{P(z_{0})}=\bar{0}=0$ .

Now conjugation respects sums and products, that is $\overline{z+w}=\bar{z}+\bar{w}$ and $\overline{zw}=\bar{z}\,\bar{w}$ . So $\overline{P(z_{0})}=\overline{a_{n}}\,\overline{z_{0}}^{\,n}+\dots+\overline{a_{1}}\,\overline{z_{0}}+\overline{a_{0}}$ . This is where the crucial hypothesis comes in: since the coefficients are REAL, they equal their own conjugates, $\overline{a_{k}}=a_{k}$ . The expression therefore becomes $a_{n}\overline{z_{0}}^{\,n}+\dots+a_{1}\overline{z_{0}}+a_{0}=P(\overline{z_{0}})$ .

Putting it together: $P(\overline{z_{0}})=0$ , so $\overline{z_{0}}$ is indeed a root. This is exactly what we see in a), where the two solutions are conjugates of each other.

c)

$P(1)=1-5+17-13=0$ , so $1$ is a root and $P$ is divisible by $(z-1)$ . Polynomial division gives $P(z)=(z-1)(z^{2}-4z+13)$ , which we check by expanding: $z^{3}-4z^{2}+13z-z^{2}+4z-13=z^{3}-5z^{2}+17z-13$ , as required.

The quadratic factor is exactly the one from part a), whose roots are $2\pm 3i$ . The three roots of $P$ are therefore $1$ , $2+3i$ and $2-3i$ .

We see the mechanism of b) at work: $P$ has degree $3$ and real coefficients, so it has at least one real root (here $1$ ), and its non-real roots necessarily come as a conjugate pair. A polynomial of odd degree with real coefficients can never have all its roots non-real, since they pair up two by two and an odd number cannot be split into pairs.

d)

For the modulus, use the exponential form. If $z=r_{1}e^{i\theta_{1}}$ and $w=r_{2}e^{i\theta_{2}}$ , then $zw=r_{1}r_{2}e^{i(\theta_{1}+\theta_{2})}$ , whose modulus is $r_{1}r_{2}=|z|\,|w|$ since the moduli are positive. In algebraic form, with $z=a+bi$ and $w=c+di$ , one can also check directly that $|zw|^{2}=(ac-bd)^{2}+(ad+bc)^{2}=a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}=(a^{2}+b^{2})(c^{2}+d^{2})=|z|^{2}|w|^{2}$ , the cross terms cancelling in pairs.

For the conjugate of the sum, with $z=a+bi$ and $w=c+di$ : $z+w=(a+c)+(b+d)i$ , so $\overline{z+w}=(a+c)-(b+d)i$ . On the other hand $\bar{z}+\bar{w}=(a-bi)+(c-di)=(a+c)-(b+d)i$ . The two expressions agree.

Finally, the counterexample. Consider $Q(z)=z-i$ , whose constant coefficient $-i$ is NOT real. Its only root is $z_{0}=i$ . But $\overline{z_{0}}=-i$ gives $Q(-i)=-i-i=-2i\neq 0$ : the conjugate is not a root. The property of b) collapses as soon as a single coefficient stops being real, because the step $\overline{a_{k}}=a_{k}$ is precisely the one carrying the whole proof. That is why the statement insists on real coefficients, and exams often penalize applying the property blindly to a polynomial with complex coefficients.

Exercise 7: Problem: the six sixth roots of 64 and the regular hexagon

We want to solve the equation $z^{6}=64$ in $\mathbb{C}$ , then use the geometric structure of its solutions.

The equation looks harmless: many students answer $z=2$ and move on. Yet degree $6$ forces six solutions, four of which are not real.

a) Write $64$ in exponential form, then find the six solutions of the equation in exponential form.
b) Give the six solutions in algebraic form.
c) Describe the figure formed by the six solutions in the complex plane. Which ones are real?
d) Compute the sum of the six solutions, and justify the result in two ways, one geometric and one algebraic.
e) Compute the product of the six solutions by grouping the conjugate roots.
f) Show that multiplication by the solution $z_{1}=1+i\sqrt{3}$ corresponds, in the plane, to a rotation combined with a dilation. Give the $2\times 2$ matrix of this transformation and check that it agrees with the modulus and the argument of $z_{1}$ .

Show the solution

Answers

a) $z_k=2e^{i\frac{k\pi}{3}}$
b) $2$ , $1\pm i\sqrt3$ , $-1\pm i\sqrt3$ , $-2$
c) Regular hexagon; $2$ and $-2$ are real
d) Zero sum
e) Product $-64$
f) $M=2R_{60^\circ}$ , $\det M=4$

a)

The number $64$ is a positive real, so its modulus is $64$ and its argument $0$ : $64=64e^{i0}$ . The sixth roots have modulus $64^{1/6}=2$ , since $2^{6}=64$ , and arguments $\dfrac{0+2k\pi}{6}=\dfrac{k\pi}{3}$ for $k=0, 1, 2, 3, 4, 5$ .

The six solutions are therefore $z_{k}=2e^{i\frac{k\pi}{3}}$ , namely $2e^{i0}$ , $2e^{i\frac{\pi}{3}}$ , $2e^{i\frac{2\pi}{3}}$ , $2e^{i\pi}$ , $2e^{i\frac{4\pi}{3}}$ and $2e^{i\frac{5\pi}{3}}$ .

b)

Convert each one using the special values of the unit circle. $z_{0}=2(1+0i)=2$ . $z_{1}=2\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right)=1+i\sqrt{3}$ . $z_{2}=2\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right)=-1+i\sqrt{3}$ . $z_{3}=2(-1+0i)=-2$ . $z_{4}=2\left(-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\right)=-1-i\sqrt{3}$ . $z_{5}=2\left(\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\right)=1-i\sqrt{3}$ .

Check on $z_{1}$ : its modulus is $2$ and its argument $\dfrac{\pi}{3}$ , so $z_{1}^{6}=2^{6}e^{i2\pi}=64$ , as required.

c)

The six solutions all have the same modulus $2$ , so they lie on the circle of radius $2$ centred at the origin, and their arguments are spaced by $\dfrac{\pi}{3}$ , that is $60^{\circ}$ . They are therefore the vertices of a REGULAR HEXAGON inscribed in that circle.

Only two of them are real: $z_{0}=2$ and $z_{3}=-2$ , those whose argument is $0$ or $\pi$ . The other four come in conjugate pairs, $z_{1}$ with $z_{5}$ and $z_{2}$ with $z_{4}$ , which is expected since the polynomial $z^{6}-64$ has real coefficients. The naive answer $z=2$ gave only one solution out of six.

d)

Sum: $2+(1+i\sqrt{3})+(-1+i\sqrt{3})+(-2)+(-1-i\sqrt{3})+(1-i\sqrt{3})$ . The real parts give $2+1-1-2-1+1=0$ and the imaginary parts $\sqrt{3}+\sqrt{3}-\sqrt{3}-\sqrt{3}=0$ . The sum is $0$ .

Geometric justification: the six points are the vertices of a regular hexagon centred at the origin. The sum of vectors arranged with point symmetry cancels out, because each vertex has its opposite in the figure ( $z_{0}$ with $z_{3}$ , $z_{1}$ with $z_{4}$ , $z_{2}$ with $z_{5}$ ). Adding three pairs of opposite vectors term by term gives the zero vector.

Algebraic justification: the six solutions are the roots of $z^{6}-64=0$ . This polynomial reads $z^{6}+0z^{5}+0z^{4}+0z^{3}+0z^{2}+0z-64$ , and by the relations between coefficients and roots, the sum of the roots is minus the coefficient of $z^{5}$ divided by the leading coefficient, that is $-\dfrac{0}{1}=0$ .

e)

Group each root with its conjugate, which makes each partial product real. The two real roots: $z_{0}z_{3}=(2)(-2)=-4$ . The pair $z_{1}z_{5}=(1+i\sqrt{3})(1-i\sqrt{3})=1^{2}+(\sqrt{3})^{2}=1+3=4$ , using $z\bar{z}=|z|^{2}$ . The pair $z_{2}z_{4}=(-1+i\sqrt{3})(-1-i\sqrt{3})=1+3=4$ in the same way.

The total product is $(-4)(4)(4)=-64$ . Check with the relations between coefficients and roots: the product of the roots of a polynomial of degree $6$ is $(-1)^{6}\dfrac{a_{0}}{a_{6}}=\dfrac{-64}{1}=-64$ . It agrees. Grouping conjugates is the right strategy: each pair immediately gives the square of the modulus, and you never handle a non-real number.

f)

Multiplying a complex number $x+iy$ by $z_{1}=1+i\sqrt{3}$ gives $(x+iy)(1+i\sqrt{3})=x+ix\sqrt{3}+iy+i^{2}y\sqrt{3}=(x-\sqrt{3}y)+i(\sqrt{3}x+y)$ . In terms of the coordinate pair $(x, y)$ , the transformation sends $(x, y)$ to $(x-\sqrt{3}y, \sqrt{3}x+y)$ , which corresponds to the matrix $M=\begin{pmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{pmatrix}$ .

Factor out the modulus: $M=2\begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\[4pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix}=2\begin{pmatrix} \cos\dfrac{\pi}{3} & -\sin\dfrac{\pi}{3} \\[4pt] \sin\dfrac{\pi}{3} & \cos\dfrac{\pi}{3} \end{pmatrix}$ . It is therefore the dilation by a factor of $2$ combined with the rotation by $\dfrac{\pi}{3}$ , that is $60^{\circ}$ .

The agreement is perfect: $|z_{1}|=2$ gives the dilation factor and $\arg(z_{1})=\dfrac{\pi}{3}$ gives the angle of the rotation. Check with the determinant: $\det(M)=(1)(1)-(-\sqrt{3})(\sqrt{3})=1+3=4=|z_{1}|^{2}$ , which is indeed the area factor expected from a dilation by a factor of $2$ .

A last reading of the whole problem: applying this transformation six times amounts to a rotation by $6\times 60^{\circ}=360^{\circ}$ , that is the identity, combined with a dilation by a factor of $2^{6}=64$ . This is exactly the statement $z_{1}^{6}=64$ , translated into the language of transformations. The hexagon of roots is nothing but the orbit of the point $(2, 0)$ under successive rotations by $60^{\circ}$ .

Exercise 8: Complex multiplication as a rotation and the equilateral triangle criterion

Multiplying by a complex number of modulus $r$ and argument $\theta$ amounts to combining a rotation by the angle $\theta$ with a dilation by a factor of $r$ . We use this geometric reading to move points of the plane.

a) Find the image of the point $z=2$ under the rotation about the origin by $60^{\circ}$ , by multiplying by the appropriate complex number.
b) Find the complex number $z'$ representing the image of the point $z=4+i$ under the rotation by $90^{\circ}$ (counterclockwise) about the centre $c=1+i$ . Recall that the formula is $z'=c+e^{i\theta}(z-c)$ .
c) Consider the points $A$ , $B$ , $C$ represented by the complex numbers $a=0$ , $b=2$ and $c=1+\sqrt{3}\,i$ . Show that triangle $ABC$ is equilateral, first by the lengths.
d) Check the algebraic criterion for an equilateral triangle, $a^{2}+b^{2}+c^{2}=ab+bc+ca$ , on this triangle, and explain the link with the rotation by $60^{\circ}$ .

Show the solution

Answers

a) $1+\sqrt3\,i$
b) $1+4i$
c) Three sides of length 2
d) $2+2\sqrt3\,i$ on both sides

a)

A rotation by $60^{\circ}$ about the origin is multiplication by $e^{i60^{\circ}}=\cos 60^{\circ}+i\sin 60^{\circ}=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$ . The image of $z=2$ is therefore $2\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right)=1+\sqrt{3}\,i$ , that is the point $(1, \sqrt{3})$ . Check: its modulus is $\sqrt{1+3}=2$ , the same as that of $z$ , which is expected since a rotation preserves distances to the origin.

b)

Apply $z'=c+e^{i90^{\circ}}(z-c)$ with $e^{i90^{\circ}}=i$ . First compute $z-c=(4+i)-(1+i)=3$ . Then $i\cdot 3=3i$ . Finally $z'=c+3i=(1+i)+3i=1+4i$ . The image is the point $1+4i$ , that is $(1, 4)$ . Translation by $-c$ , rotation, then translation by $+c$ : this is the universal pattern of a rotation about a centre other than the origin.

c)

Compute the three lengths as moduli of differences of the complex numbers. $AB=|b-a|=|2|=2$ . $AC=|c-a|=|1+\sqrt{3}\,i|=\sqrt{1+3}=2$ . $BC=|c-b|=|(1+\sqrt{3}\,i)-2|=|-1+\sqrt{3}\,i|=\sqrt{1+3}=2$ . The three sides all equal $2$ , so triangle $ABC$ is equilateral.

d)

Check $a^{2}+b^{2}+c^{2}=ab+bc+ca$ . On the left: $0+4+(1+\sqrt{3}\,i)^{2}=4+(1+2\sqrt{3}\,i-3)=4+(-2+2\sqrt{3}\,i)=2+2\sqrt{3}\,i$ . On the right: $ab+bc+ca=0+2(1+\sqrt{3}\,i)+0=2+2\sqrt{3}\,i$ . The two sides agree, the criterion holds. The link with the rotation: a triangle $ABC$ is equilateral and positively oriented if and only if $C$ is the image of $B$ under the rotation by $60^{\circ}$ about $A$ , which reads $c-a=e^{i60^{\circ}}(b-a)$ . Expanding this relation, with $e^{i60^{\circ}}$ a root of $t^{2}-t+1=0$ , gives precisely the symmetric identity $a^{2}+b^{2}+c^{2}=ab+bc+ca$ . The geometry of the equilateral triangle is thus encoded in a single equation between the complex numbers $a$ , $b$ and $c$ .

Exercise 9: De Moivre's theorem and the triple-angle formulas

De Moivre's theorem is not only for computing powers: once expanded, it produces the multiple-angle trigonometric formulas. Here we apply it to the triple angle.

a) Write $(\cos\theta+i\sin\theta)^{3}$ in two ways: by De Moivre's theorem, and by expanding the cube of a sum.
b) By identifying the real and imaginary parts, deduce the formulas for $\cos 3\theta$ and $\sin 3\theta$ in terms of $\cos\theta$ and $\sin\theta$ .
c) Express $\cos 3\theta$ in terms of $\cos\theta$ only, then $\sin 3\theta$ in terms of $\sin\theta$ only.
d) Check the formula for $\cos 3\theta$ at $\theta=30^{\circ}$ , where $\cos 3\theta=\cos 90^{\circ}=0$ .

Show the solution

Answers

a) $\cos 3\theta+i\sin 3\theta=(c^3-3cs^2)+i(3c^2s-s^3)$
b) $\cos 3\theta=c^3-3cs^2$ , $\sin 3\theta=3c^2s-s^3$
c) $4\cos^3\theta-3\cos\theta$ ; $3\sin\theta-4\sin^3\theta$
d) $0$ for $\theta=30^\circ$

a)

By De Moivre's theorem: $(\cos\theta+i\sin\theta)^{3}=\cos 3\theta+i\sin 3\theta$ . By expanding the cube, with $c=\cos\theta$ and $s=\sin\theta$ : $(c+is)^{3}=c^{3}+3c^{2}(is)+3c(is)^{2}+(is)^{3}=c^{3}+3ic^{2}s-3cs^{2}-is^{3}$ , using $i^{2}=-1$ and $i^{3}=-i$ . Collecting: $(c+is)^{3}=(c^{3}-3cs^{2})+i(3c^{2}s-s^{3})$ .

b)

Identify the two expressions. Real part: $\cos 3\theta=\cos^{3}\theta-3\cos\theta\sin^{2}\theta$ . Imaginary part: $\sin 3\theta=3\cos^{2}\theta\sin\theta-\sin^{3}\theta$ . This is the fundamental mechanism: a single application of De Moivre's theorem delivers both triple-angle formulas at once, where classical trigonometry needs the addition formula twice.

c)

For $\cos 3\theta$ , replace $\sin^{2}\theta=1-\cos^{2}\theta$ : $\cos 3\theta=\cos^{3}\theta-3\cos\theta(1-\cos^{2}\theta)=\cos^{3}\theta-3\cos\theta+3\cos^{3}\theta=4\cos^{3}\theta-3\cos\theta$ . For $\sin 3\theta$ , replace $\cos^{2}\theta=1-\sin^{2}\theta$ : $\sin 3\theta=3(1-\sin^{2}\theta)\sin\theta-\sin^{3}\theta=3\sin\theta-3\sin^{3}\theta-\sin^{3}\theta=3\sin\theta-4\sin^{3}\theta$ . These two polynomials, in $\cos\theta$ and in $\sin\theta$ , are special cases of the Chebyshev polynomials.

d)

For $\theta=30^{\circ}$ , $\cos\theta=\dfrac{\sqrt{3}}{2}$ . The formula gives $\cos 90^{\circ}=4\left(\dfrac{\sqrt{3}}{2}\right)^{3}-3\cdot\dfrac{\sqrt{3}}{2}=4\cdot\dfrac{3\sqrt{3}}{8}-\dfrac{3\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}-\dfrac{3\sqrt{3}}{2}=0$ . We do get $\cos 90^{\circ}=0$ , which validates the formula on a known value.

Exercise 10: Problem: loci in the complex plane

The expression $|z-a|$ reads as the distance from the point $z$ to the point $a$ in the complex plane. This single reading turns complex equations into classic geometric figures: lines, circles, perpendicular bisectors.

a) Find and describe the set of points $z$ such that $|z-1|=|z+i|$ . Give its Cartesian equation.
b) Find the set of points $z$ such that $|z-(2+i)|=3$ . State its nature, its centre and its radius, and say whether the point $4+2i$ lies on it, inside it or outside it.
c) Find the set of points $z$ such that $|z-1|=2\,|z+1|$ . Show that it is a circle and give its centre and its radius.
d) Describe geometrically the set of points $z=x+iy$ such that $\text{Re}(z)+\text{Im}(z)=1$ , and explain why, unlike b) and c), this locus is a line.

Show the solution

Answers

a) Line $y=-x$
b) Circle with centre $(2, 1)$ , radius 3; point inside
c) Circle with centre $\left(-\frac53, 0\right)$ , radius $\frac43$
d) Line $x+y=1$

a)

Setting $z=x+iy$ : $|z-1|^{2}=(x-1)^{2}+y^{2}$ and $|z+i|^{2}=x^{2}+(y+1)^{2}$ . The equality $|z-1|=|z+i|$ is equivalent to $(x-1)^{2}+y^{2}=x^{2}+(y+1)^{2}$ , that is $x^{2}-2x+1+y^{2}=x^{2}+y^{2}+2y+1$ , hence $-2x=2y$ and $y=-x$ . The set is the LINE with equation $y=-x$ . Interpretation: $|z-1|=|z+i|$ means that $z$ is equidistant from the points $1$ and $-i$ , so it is the perpendicular bisector of the segment joining these two points.

b)

$|z-(2+i)|=3$ means that the distance from $z$ to the point $2+i$ is always $3$ : it is the CIRCLE with centre $(2, 1)$ and radius $3$ . For the point $4+2i$ , compute its distance to the centre: $|(4+2i)-(2+i)|=|2+i|=\sqrt{4+1}=\sqrt{5}\approx 2.24$ . Since $\sqrt{5}<3$ , this point is INSIDE the disk, it is not on the circle.

c)

Setting $z=x+iy$ and squaring: $(x-1)^{2}+y^{2}=4\big((x+1)^{2}+y^{2}\big)$ . Expand: $x^{2}-2x+1+y^{2}=4x^{2}+8x+4+4y^{2}$ , that is $0=3x^{2}+3y^{2}+10x+3$ , hence $x^{2}+y^{2}+\dfrac{10}{3}x+1=0$ . Complete the square: $\left(x+\dfrac{5}{3}\right)^{2}+y^{2}=\dfrac{25}{9}-1=\dfrac{16}{9}$ . It is the CIRCLE with centre $\left(-\dfrac{5}{3}, 0\right)$ and radius $\dfrac{4}{3}$ . This is a circle of Apollonius: the locus of points whose distances to two fixed points have a constant ratio different from $1$ is always a circle.

d)

The condition $\text{Re}(z)+\text{Im}(z)=1$ reads directly $x+y=1$ : it is a LINE. The difference with b) and c) lies in the nature of the equation. A condition on the real and imaginary PARTS of $z$ is linear in $x$ and $y$ , so it gives a line. A condition of EQUAL MODULI, as in a), produces squares that cancel and also leaves a linear equation (a perpendicular bisector). But a condition of CONSTANT modulus, as in b), or of a RATIO of moduli different from $1$ , as in c), keeps the terms in $x^{2}+y^{2}$ and produces a circle. Recognizing the type of equation lets you predict the figure before calculating anything.

Exercise 11: The fifth roots of unity and the cosine of 72°

Let $\omega=e^{i\frac{2\pi}{5}}$ .

a) Give the solutions of $z^{5}=1$ and describe the figure they form.
b) Show that $1+\omega+\omega^{2}+\omega^{3}+\omega^{4}=0$ .
c) Let $t=\omega+\omega^{4}$ . Show that $t=2\cos\frac{2\pi}{5}$ , that $\omega^{2}+\omega^{3}=t^{2}-2$ , then that $t^{2}+t-1=0$ .
d) Deduce the exact value of $\cos 72^{\circ}$ , then its value to three decimal places.
e) Compute the side length of the regular pentagon inscribed in the unit circle, to three decimal places.

Show the solution

Answers

a) Regular pentagon
b) Geometric sum equal to zero
c) $t^2+t-1=0$
d) $\cos 72^\circ=\frac{\sqrt5-1}{4}\approx 0.309$
e) Side $\approx 1.176$

a)

$z_{k}=e^{i\frac{2k\pi}{5}}$ for $k=0, 1, 2, 3, 4$ , that is $1$ , $\omega$ , $\omega^{2}$ , $\omega^{3}$ , $\omega^{4}$ . Five points of the unit circle spaced by $72^{\circ}$ : a REGULAR PENTAGON.

b)

It is a geometric sum with ratio $\omega\neq 1$ : $\frac{\omega^{5}-1}{\omega-1}=\frac{1-1}{\omega-1}=0$ . Geometrically, the vertices of a regular polygon centred at the origin add up to zero.

c)

$\omega^{4}=e^{-i\frac{2\pi}{5}}=\bar{\omega}$ , so $t=\omega+\bar{\omega}=2\cos\frac{2\pi}{5}$ . Then $t^{2}=\omega^{2}+2\omega^{5}+\omega^{8}=\omega^{2}+2+\omega^{3}$ , since $\omega^{5}=1$ and $\omega^{8}=\omega^{3}$ : $\omega^{2}+\omega^{3}=t^{2}-2$ . By b), $1+t+(t^{2}-2)=0$ , that is $t^{2}+t-1=0$ .

d)

$t=\frac{-1\pm\sqrt{5}}{2}$ , and $t=2\cos 72^{\circ}>0$ forces $t=\frac{\sqrt{5}-1}{2}$ . So $\cos 72^{\circ}=\frac{\sqrt{5}-1}{4}\approx 0.309$ . The negative root, $\frac{-1-\sqrt{5}}{2}$ , equals $\omega^{2}+\omega^{3}=2\cos 144^{\circ}$ .

e)

The side is $|\omega-1|$ , and $|\omega-1|^{2}=(\cos 72^{\circ}-1)^{2}+\sin^{2}72^{\circ}=2-2\cos 72^{\circ}=2-\frac{\sqrt{5}-1}{2}=\frac{5-\sqrt{5}}{2}\approx 1.382$ . Side $\approx 1.176$ .

The thread: roots of unity turn a trigonometry question into a polynomial equation; their zero sum provides the key relation.

Exercise 12: The factorization of z⁴ + 4

We want to solve $z^{4}+4=0$ in $\mathbb{C}$ , then derive a real factorization from it.

a) Write $-4$ in exponential form and find the four solutions in algebraic form.
b) By grouping the conjugate roots, write $z^{4}+4$ as a product of two quadratic polynomials with real coefficients.
c) Check the factorization by noticing that $z^{4}+4=(z^{2}+2)^{2}-(2z)^{2}$ .
d) Deduce that $n^{4}+4$ is never a prime number for an integer $n\geq 2$ . Factor $3^{4}+4$ and $5^{4}+4$ .
e) Compute the sum and the product of the four solutions, and check them against the coefficients of $z^{4}+4$ .

Show the solution

Answers

a) $\pm1\pm i$
b) $(z^2-2z+2)(z^2+2z+2)$
c) Difference of two squares
d) $85=5\times 17$ ; $629=17\times 37$
e) Sum 0, product 4

a)

$-4=4e^{i\pi}$ . The fourth roots have modulus $4^{1/4}=\sqrt{2}$ and arguments $\frac{\pi}{4}+\frac{k\pi}{2}$ : $\sqrt{2}\,e^{i\frac{\pi}{4}}=1+i$ , then $-1+i$ , $-1-i$ and $1-i$ . They form a square.

b)

$(z-1-i)(z-1+i)=(z-1)^{2}+1=z^{2}-2z+2$ and $(z+1-i)(z+1+i)=(z+1)^{2}+1=z^{2}+2z+2$ . So $z^{4}+4=(z^{2}-2z+2)(z^{2}+2z+2)$ .

c)

$(z^{2}+2)^{2}-(2z)^{2}=z^{4}+4z^{2}+4-4z^{2}=z^{4}+4$ , and a difference of two squares factors as $(z^{2}+2-2z)(z^{2}+2+2z)$ : this is indeed the factorization of b).

d)

$n^{4}+4=(n^{2}-2n+2)(n^{2}+2n+2)$ , and for $n\geq 2$ the first factor equals $(n-1)^{2}+1\geq 2$ , while the second is larger still: the number has two factors greater than $1$ , so it is not prime. $3^{4}+4=85=5\times 17$ and $5^{4}+4=629=17\times 37$ .

e)

Sum: $(1+i)+(-1+i)+(-1-i)+(1-i)=0$ , and the coefficient of $z^{3}$ is zero. Product: $(1+i)(1-i)\times(-1+i)(-1-i)=2\times 2=4$ , equal to the constant term, since the degree is even.

The thread: the complex roots of a real polynomial group into conjugate pairs that give real quadratic factors.

Exercise 13: A quadratic equation with complex coefficients

We solve the equation $z^{2}-(3+i)z+4+3i=0$ in $\mathbb{C}$ .

a) Compute the discriminant $\Delta$ .
b) Find the two square roots of $\Delta$ by the algebraic method.
c) Deduce the two solutions of the equation.
d) Check them using the sum and the product of the solutions.
e) Are the two solutions conjugate? Why does this not contradict the conjugate root property?

Show the solution

Answers

a) $\Delta=-8-6i$
b) $\pm(1-3i)$
c) $2-i$ and $1+2i$
d) Sum $3+i$ , product $4+3i$
e) No: the coefficients are not real

a)

$\Delta=(3+i)^{2}-4(4+3i)=9+6i-1-16-12i=-8-6i$ .

b)

Look for $\delta=x+iy$ with $x^{2}-y^{2}=-8$ , $2xy=-6$ and $x^{2}+y^{2}=|\Delta|=\sqrt{64+36}=10$ . Hence $x^{2}=1$ and $y^{2}=9$ , and $xy<0$ forces opposite signs: $\delta=1-3i$ or $\delta=-1+3i$ .

c)

$z=\frac{(3+i)\pm(1-3i)}{2}$ : $z_{1}=\frac{4-2i}{2}=2-i$ and $z_{2}=\frac{2+4i}{2}=1+2i$ .

d)

Sum: $(2-i)+(1+2i)=3+i$ , the opposite of the coefficient of $z$ . Product: $(2-i)(1+2i)=2+4i-i-2i^{2}=4+3i$ , the constant term. Both relations hold.

e)

No: the conjugate of $2-i$ is $2+i$ , which is not a solution. There is no contradiction: the conjugate root property only holds for a polynomial with REAL coefficients, and here two coefficients are not real.

The thread: the quadratic formula remains valid in $\mathbb{C}$ ; the only new step is extracting a complex square root.

Exercise 14: Problem: the iterations of the Mandelbrot set

For a complex number $c$ , define the sequence $z_{0}=0$ and $z_{n+1}=z_{n}^{2}+c$ . The number $c$ belongs to the Mandelbrot set if this sequence stays bounded. Take as given that as soon as a term satisfies $|z_{n}|>2$ , the sequence is unbounded.

a) Compute the terms $z_{1}$ to $z_{4}$ for $c=1$ . Does the number $1$ belong to the set?
b) Same question for $c=-1$ .
c) Compute $z_{1}$ to $z_{4}$ for $c=i$ and conclude.
d) For $c=-2$ , compute the first terms. For $c=0.5$ , find the first index $n$ such that $|z_{n}|>2$ .
e) For real $c$ strictly greater than $\frac{1}{4}$ , show that $z_{n+1}-z_{n}\geq c-\frac{1}{4}$ , and conclude.

Show the solution

Answers

a) $1, 2, 5, 26$ : no
b) $-1, 0, -1, 0$ : yes
c) $i$ , $-1+i$ , $-i$ , $-1+i$ : yes
d) $c=-2$ : bounded; $c=0.5$ : $n=5$
e) Increase of at least $c-\frac14$ : divergence

a)

$z_{1}=1$ , $z_{2}=2$ , $z_{3}=5$ , $z_{4}=26$ . From $z_{3}$ on, the modulus exceeds $2$ : the sequence is unbounded, $1$ does not belong to the set.

b)

$z_{1}=-1$ , $z_{2}=0$ , $z_{3}=-1$ , $z_{4}=0$ : the sequence alternates between $-1$ and $0$ , it is bounded, and $-1$ belongs to the set.

c)

$z_{1}=i$ ; $z_{2}=i^{2}+i=-1+i$ ; $z_{3}=(-1+i)^{2}+i=(1-2i-1)+i=-i$ ; $z_{4}=(-i)^{2}+i=-1+i$ . The sequence repeats $-1+i$ , $-i$ : it is bounded, with modulus at most $\sqrt{2}$ , and $i$ belongs to the set.

d)

For $c=-2$ : $z_{1}=-2$ , $z_{2}=2$ , then $z_{n}=2$ for every $n\geq 2$ . The sequence is bounded, and $-2$ is the leftmost point of the set. For $c=0.5$ : $z_{1}=0.5$ , $z_{2}=0.75$ , $z_{3}=1.0625$ , $z_{4}\approx 1.629$ , $z_{5}\approx 3.153$ : the first index is $n=5$ .

e)

For real $c$ , every term is real, and $z_{n+1}-z_{n}=z_{n}^{2}-z_{n}+c=\left(z_{n}-\frac{1}{2}\right)^{2}+c-\frac{1}{4}\geq c-\frac{1}{4}$ . The sequence therefore increases by at least $c-\frac{1}{4}>0$ at each step: $z_{n}\geq n\left(c-\frac{1}{4}\right)$ tends to infinity. On the real axis, the set does not extend beyond $\frac{1}{4}$ on the right.

The thread: iterating $z\mapsto z^{2}+c$ combines the product and the sum of complex numbers; the modulus decides divergence.

Exercise 15: Problem: adding sinusoidal signals with phasors

A signal $a\cos(\omega t+\varphi)$ , with $a>0$ , is represented by the complex number $a\,e^{i\varphi}$ , its PHASOR: the signal is the real part of $a\,e^{i\varphi}e^{i\omega t}$ . Two signals with the same angular frequency are added by adding their phasors.

a) Give the phasors of $s_{1}(t)=3\cos(\omega t)$ and of $s_{2}(t)=4\cos\left(\omega t+\frac{\pi}{2}\right)$ .
b) Compute the phasor of $s_{1}+s_{2}$ in exponential form: amplitude and phase, to the hundredth of a degree.
c) Write $s_{1}+s_{2}$ in the form $A\cos(\omega t+\Phi)$ and check the value at $t=0$ .
d) Three signals $2\cos(\omega t)$ , $2\cos\left(\omega t+\frac{2\pi}{3}\right)$ and $2\cos\left(\omega t+\frac{4\pi}{3}\right)$ are added. What is the amplitude of the sum?
e) Differentiating $a\cos(\omega t+\varphi)$ amounts to multiplying its phasor by $i\omega$ . For $\omega=2\ \mathrm{rad/s}$ , give the phasor and the amplitude of the derivative of $s_{1}$ .

Show the solution

Answers

a) $3$ and $4i$
b) $5e^{i\,0.927}$ , that is $53.13^\circ$
c) $5\cos(\omega t+0.927)$ ; 3 at $t=0$
d) Zero sum
e) Phasor $6i$ , amplitude 6

a)

$s_{1}$ : $3e^{i0}=3$ . $s_{2}$ : $4e^{i\frac{\pi}{2}}=4i$ .

b)

$3+4i$ , with modulus $5$ and argument $\arctan\frac{4}{3}\approx 0.927\ \mathrm{rad}$ , that is $53.13^{\circ}$ : $5e^{i\,0.927}$ .

c)

$s_{1}(t)+s_{2}(t)=5\cos(\omega t+0.927)$ . At $t=0$ : $3+4\cos\frac{\pi}{2}=3$ , and $5\cos(0.927)=5\times 0.6=3$ . The amplitude of the sum, $5$ , is not $3+4=7$ : the two signals are shifted by a quarter of a period.

d)

The three phasors $2$ , $2e^{i\frac{2\pi}{3}}$ and $2e^{i\frac{4\pi}{3}}$ are the vertices of an equilateral triangle centred at the origin: their sum is zero, and the sum of the signals is identically zero. This is the principle of three-phase current, whose three phases balance out.

e)

The phasor of $s_{1}'$ is $2i\times 3=6i=6e^{i\frac{\pi}{2}}$ : $s_{1}'(t)=6\cos\left(2t+\frac{\pi}{2}\right)=-6\sin(2t)$ , with amplitude $6$ . This is indeed the derivative of $3\cos(2t)$ .

The thread: phasors replace the trigonometry of signals with the same frequency by the addition and multiplication of complex numbers.

Corrected exercises: complex numbers, De Moivre's theorem and nth roots (201-SN4-RE)

Course recap

Calculations: working in algebraic form

Exercise 0.1: Computing in algebraic form: ten direct calculations

Calculations: exponential form and De Moivre's theorem

Exercise 0.2: Converting to exponential form and computing powers: nine direct calculations

Part A: The basics (/50)

Exercise 1: Algebraic form, conjugate and division

Exercise 2: Modulus, argument and exponential form

Exercise 3: De Moivre's theorem and powers

Exercise 4: The nth roots of a complex number

Exercise 5: Square roots of a complex number by the algebraic method

Part B: Exam level (/50)

Exercise 6: Polynomial equations and conjugate roots

Exercise 7: Problem: the six sixth roots of 64 and the regular hexagon

Exercise 8: Complex multiplication as a rotation and the equilateral triangle criterion

Exercise 9: De Moivre's theorem and the triple-angle formulas

Exercise 10: Problem: loci in the complex plane

Part C: The classics (/50)

Exercise 11: The fifth roots of unity and the cosine of 72°

Exercise 12: The factorization of z⁴ + 4

Exercise 13: A quadratic equation with complex coefficients

Exercise 14: Problem: the iterations of the Mandelbrot set

Exercise 15: Problem: adding sinusoidal signals with phasors

See also

Looking for a linear algebra tutor in Montreal?