Linear Algebra and Vector Geometry 201-SN4-RE • CEGEP, Montreal

Revision sheet: complex numbers and De Moivre's theorem (201-SN4-RE)

Q: How do you find the argument of a complex number?

First compute the modulus, then write two equations: the cosine equals the real part divided by the modulus, and the sine equals the imaginary part divided by the modulus. It is the signs of these two quantities together that fix the quadrant. The tangent alone is never enough, because it cannot tell a number from its opposite.

Q: Why does the equation z to the power n equals a have n solutions?

Because the argument of a complex number is only defined up to a multiple of two pi. By writing the right-hand side with this periodic term before taking the root, you get n different angles, all within one full turn. The solutions are then spread evenly around a circle, at the vertices of a regular polygon.

Q: How do you divide two complex numbers?

Multiply the numerator and the denominator by the conjugate of the denominator. The denominator then becomes the square of the modulus, a real number, and the quotient can be written in algebraic form. In exponential form it is even simpler: the moduli divide and the arguments subtract.

Q: When should you use De Moivre's theorem?

Whenever you need to raise a complex number to a power greater than two, or to solve an equation where the unknown is raised to a power. Put the number in exponential form, raise the modulus to the power and multiply the argument by the exponent. Expanding in algebraic form would be much longer and much riskier.

Q: Is the modulus of a sum the sum of the moduli?

No, except in a special case. The modulus behaves like a length: it distributes over a product, but not over a sum. There is only an inequality, the modulus of the sum being less than or equal to the sum of the moduli, with equality only when the two numbers have the same argument.

Q: Can you write the square root of a negative number?

Better not. Over the complex numbers, every non-zero number has two square roots, so the usual notation loses its single meaning. Write the result directly with i, for example two i for minus four. The rule for multiplying two square roots, by the way, only holds for positive numbers.

This sheet does not repeat the formulas for switching between the three forms: they are in your notes. It deals with what decides the mark, namely choosing the form according to the operation, finding the argument correctly, and the write-up expected for an equation $z^{n}=a$ .

It is written for CEGEP students in the Science program taking 201-SN4-RE (formerly 201-NYC). Once you have read it, the corrected exercise set for the same chapter puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The algebraic form is for ADDING, the exponential form for MULTIPLYING and raising to a power. Almost every lost mark comes from keeping the wrong form, or from choosing the argument with the cosine alone.

This chapter is part of 201-SN4-RE, Linear Algebra and Vector Geometry

The essentials

Three forms, and what each one is for

• ALGEBRAIC FORM $z=a+bi$ , with $i^{2}=-1$ : this is the form for ADDITION and subtraction, which are done component by component.
• Conjugate $\bar{z}=a-bi$ , modulus $|z|=\sqrt{a^{2}+b^{2}}$ , and the key relation $z\bar{z}=a^{2}+b^{2}=|z|^{2}$ , which is a positive REAL number.
• TRIGONOMETRIC FORM $z=r(\cos\theta+i\sin\theta)$ and EXPONENTIAL FORM $z=r\,e^{i\theta}$ , with $r=|z|$ and $\theta=\arg(z)$ .
• These are the forms for PRODUCTS, quotients and powers: $r_{1}e^{i\theta_{1}}\times r_{2}e^{i\theta_{2}}=r_{1}r_{2}e^{i(\theta_{1}+\theta_{2})}$ . The moduli multiply, the arguments add.
• DIVISION in algebraic form: multiply the top and the bottom by the conjugate of the denominator, which makes it real: $\dfrac{z_{1}}{z_{2}}=\dfrac{z_{1}\bar{z_{2}}}{|z_{2}|^{2}}$ .

For

z=3+2i

: the modulus

|z|

is the length of the segment from the origin to

z

, the argument

\theta

its angle with the real axis, the conjugate

\bar{z}=3-2i

its mirror image in that axis.

The first reflex in front of an exercise: look at the OPERATION asked for. A sum stays algebraic, a power goes exponential, and switching forms costs less than sticking with the wrong one.

The argument is fixed by a PAIR, never by a single ratio

• Solve $\cos\theta=\dfrac{a}{r}$ AND $\sin\theta=\dfrac{b}{r}$ : it is the two equations together that fix the quadrant.
• The tangent alone is never enough: $z$ and $-z$ have exactly the same ratio $\frac{b}{a}$ , hence the same tangent, while their arguments differ by $\pi$ .
• The sign of $a$ gives one half of the plane, the sign of $b$ the other: positive and positive for the first quadrant, negative and positive for the second, and so on.
• The argument is defined MODULO $2\pi$ : $\frac{\pi}{3}$ and $\frac{7\pi}{3}$ are the same argument, and the question often asks for the one in $(-\pi, \pi]$ .
• Special cases to know without calculation: $\arg(1)=0$ , $\arg(i)=\frac{\pi}{2}$ , $\arg(-1)=\pi$ , $\arg(-i)=-\frac{\pi}{2}$ .

For

z=-\sqrt{3}+i

, the arctangent gives

-30^{\circ}

(in red), where

z

is not. The pair

\cos\theta<0

\sin\theta>0

puts

z

in the second quadrant:

\theta=150^{\circ}

(in green).

Sketching the point in the complex plane before concluding takes ten seconds and makes a quadrant error impossible. It is the most profitable check of the chapter.

De Moivre's theorem, the roots and the regular polygon

• DE MOIVRE'S THEOREM: $\left(re^{i\theta}\right)^{n}=r^{n}e^{in\theta}$ , that is $r^{n}\left(\cos(n\theta)+i\sin(n\theta)\right)$ . The modulus is raised to the power, the argument is MULTIPLIED.
• The $n$ TH ROOTS of $z=re^{i\theta}$ : there are exactly $n$ of them, given by $z_{k}=r^{1/n}\,e^{i\left(\frac{\theta+2k\pi}{n}\right)}$ for $k=0, 1, \dots, n-1$ .
• The term $2k\pi$ is not decoration: it is what produces the $n$ distinct roots. Without it, you find only one.
• These roots lie on the circle of radius $r^{1/n}$ , evenly spaced by $\frac{2\pi}{n}$ : they form a REGULAR POLYGON with $n$ sides, and their sum is $0$ .
• CONJUGATE ROOTS: if a polynomial has REAL coefficients and has $z_{0}$ as a root, then $\bar{z_{0}}$ is a root too. Non-real roots come in pairs.

The three cube roots of

8

: modulus

2

, arguments

0

\frac{2\pi}{3}

\frac{4\pi}{3}

. Spaced by

120^{\circ}

on the circle of radius

2

, they form an equilateral triangle.

Once the first root is found, the others are obtained by adding $\frac{2\pi}{n}$ to the argument, without recomputing the modulus. That is what lets you handle the six sixth roots in three lines.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Taking the arctangent without looking at the quadrant

the whole question, and the error carries over to every power computed afterwards.

What not to write

“ $z=-1-i$ gives $\tan\theta=\frac{-1}{-1}=1$ , so $\theta=\frac{\pi}{4}$ .”

What to write

“ $\cos\theta=\frac{-1}{\sqrt{2}}$ and $\sin\theta=\frac{-1}{\sqrt{2}}$ : both are negative, so $z$ is in the THIRD quadrant and $\theta=\frac{5\pi}{4}$ , or $-\frac{3\pi}{4}$ .”

z=1+i

and

-z=-1-i

lie on the SAME line through the origin, so they have the same tangent,

1

; yet their arguments differ by

180^{\circ}

45^{\circ}

and

225^{\circ}

Why: The tangent cannot tell $z$ from $-z$ : they have the same ratio $\frac{b}{a}$ . Only the cosine-sine PAIR fixes the quadrant, and that is why the course writes both equations.

2. Dividing without using the conjugate

2 marks, and a wrong sign that contaminates the rest of the exercise.

What not to write

“ $\dfrac{1}{i}=i$ , since $i\times i$ does give something simple.”

What to write

“ $\dfrac{1}{i}=\dfrac{\bar{i}}{i\bar{i}}=\dfrac{-i}{|i|^{2}}=\dfrac{-i}{1}=-i$ . Multiply the top and the bottom by the conjugate of the denominator, which makes it REAL.”

Why: A quotient of complex numbers has no algebraic form until the denominator is real. The conjugate is the only tool that achieves this, thanks to $z\bar{z}=|z|^{2}$ .

3. Distributing the modulus over a sum

the whole proof question, and a wrong numerical result from the first calculation.

What not to write

“ $|z_{1}+z_{2}|=|z_{1}|+|z_{2}|$ .”

What to write

“The modulus distributes over a PRODUCT, not over a sum: $|z_{1}z_{2}|=|z_{1}|\,|z_{2}|$ , but $|z_{1}+z_{2}|\leq|z_{1}|+|z_{2}|$ , with equality only if the two have the same argument.”

The direct path from the origin to the point

z_{1}+z_{2}

is shorter than the detour through

z_{1}

: this is exactly the inequality

|z_{1}+z_{2}|\leq|z_{1}|+|z_{2}|

Why: The modulus is a length: this is the ordinary triangle inequality. Two vectors add up in length only when they point in the same direction.

4. Applying De Moivre's theorem to the algebraic form

the whole question, and the answer is off by a factor of

8

What not to write

“ $(1+i)^{8}=1^{8}+i^{8}=1+1=2$ .”

What to write

“De Moivre's theorem applies to the trigonometric form. $1+i=\sqrt{2}\,e^{i\pi/4}$ , so $(1+i)^{8}=\left(\sqrt{2}\right)^{8}e^{2i\pi}=16$ .”

Why: A power never distributes over a sum, no more for complex numbers than for reals. Switching to exponential form is the only way to turn the power into a multiplication of the angle.

5. Giving only one nth root

five sixths of the question, and it is the most frequent error of the chapter.

What not to write

“ $z^{6}=64$ gives $z=64^{1/6}=2$ .”

What to write

“There are SIX roots: $z_{k}=2\,e^{i\frac{2k\pi}{6}}$ for $k=0$ to $5$ , namely $2$ , $1+i\sqrt{3}$ , $-1+i\sqrt{3}$ , $-2$ , $-1-i\sqrt{3}$ and $1-i\sqrt{3}$ .”

The six roots of

z^{6}=64

lie on the circle of radius

2

, spaced by

60^{\circ}

: a regular hexagon. Answering

z=2

alone gives only

z_{0}

, in green.

Why: The argument of a complex number is defined modulo $2\pi$ : writing $64=64e^{i(0+2k\pi)}$ before taking the root is what brings out the $n$ solutions. Without the $2k\pi$ , you see only one.

6. Multiplying two square roots of negative numbers

the whole question, with a reversed sign that shows up nowhere else.

What not to write

“ $\sqrt{-4}\times\sqrt{-9}=\sqrt{36}=6$ .”

What to write

“The rule $\sqrt{a}\sqrt{b}=\sqrt{ab}$ only holds for $a\geq 0$ and $b\geq 0$ . Here $\sqrt{-4}=2i$ and $\sqrt{-9}=3i$ , so the product is $6i^{2}=-6$ .”

Why: The notation $\sqrt{\ }$ no longer has a single meaning over the complex numbers: every non-zero number has TWO square roots. So write $2i$ from the start, and never keep a radical of a negative number.

7. Invoking conjugate roots for a polynomial with complex coefficients

2 marks, and an invented root that ruins the whole factorization.

What not to write

“ $z_{0}=1+2i$ is a root, so $1-2i$ is a root too, whatever the polynomial.”

What to write

“This property requires REAL coefficients. The polynomial $z-(1+2i)$ has $1+2i$ as its only root, and $1-2i$ is not one: check the coefficients before invoking the theorem.”

Why: The proof rests on $\overline{P(z)}=P(\bar{z})$ , an equality that only holds if every coefficient equals its own conjugate, that is, if they are real.

8. Confusing $z\bar{z}$ with the modulus

1 mark each time, and one division in two comes out wrong.

What not to write

“ $z\bar{z}=|z|$ .”

What to write

“ $z\bar{z}=|z|^{2}$ , with the SQUARE: for $z=3+4i$ , $z\bar{z}=9+16=25$ while $|z|=5$ .”

Why: $z\bar{z}=(a+bi)(a-bi)=a^{2}+b^{2}$ , which is the square of the root that defines the modulus. That is also why the denominator of a division is $|z_{2}|^{2}$ and not $|z_{2}|$ .

Which method to choose

Which form to choose, depending on the operation

Look at the OPERATION before the numbers. Each operation has its form, and switching forms costs less than persisting.

If sum or difference → ALGEBRAIC form, component by component

Example: $(2+3i)+(1-i)=3+2i$
If product or quotient of two numbers given in algebraic form → stay algebraic if the numbers are simple, expanding and using the conjugate to divide

Beyond two factors, the exponential form becomes faster.
If power with exponent $3$ or more → EXPONENTIAL form and De Moivre's theorem

Example: $(1+i)^{8}=\left(\sqrt{2}e^{i\pi/4}\right)^{8}=16$
If equation of the type $z^{n}=a$ → exponential form of the right-hand side, with the $2k\pi$ , then $n$ roots

Example: $z^{6}=64$ gives $z_{k}=2e^{i\frac{2k\pi}{6}}$
If locus, or a condition on a modulus → set $z=x+iy$ and translate into a Cartesian equation

$|z-z_{0}|=R$ is the circle with centre $z_{0}$ and radius $R$ , with no calculation at all.
If polynomial equation with real coefficients → algebraic form and discriminant, remembering that non-real roots come in conjugate pairs

No branch applies? Look at what the ANSWER must be: a locus is described in Cartesian form, an exact number often in exponential form, a numerical value in algebraic form.

How to fix the argument, quadrant by quadrant

First compute $r=|z|$ , then read the SIGNS of $a$ and $b$ before looking for an angle.

The signs of

a

and

b

give the quadrant before any calculation:

1+i

-1+i

-1-i

and

1-i

all have

|\tan\theta|=1

, but four different arguments, one per quadrant.

If $a>0$ and $b>0$ → first quadrant: $\theta\in\left(0, \frac{\pi}{2}\right)$ , and the arctangent gives the answer directly

Example: $1+i$ gives $\frac{\pi}{4}$
If $a<0$ and $b>0$ → second quadrant: add $\pi$ to the arctangent

Example: $-1+i$ gives $\frac{3\pi}{4}$
If $a<0$ and $b<0$ → third quadrant: add $\pi$ to the arctangent, or subtract $\pi$

Example: $-1-i$ gives $\frac{5\pi}{4}$ , or $-\frac{3\pi}{4}$
If $a>0$ and $b<0$ → fourth quadrant: the arctangent gives a negative angle, which is correct as it stands

Example: $1-i$ gives $-\frac{\pi}{4}$
If $a=0$ or $b=0$ → special case to read directly: $0$ , $\frac{\pi}{2}$ , $\pi$ or $-\frac{\pi}{2}$

The arctangent is not even defined when $a=0$ : this is where the formula fails most visibly.

The final check, valid in every case: recompute $r\cos\theta$ and $r\sin\theta$ and make sure you get $a$ and $b$ back, SIGNS included.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Putting a complex number in exponential form

When to use it: The question asks for the trigonometric or exponential form, or involves a high power.

1 Compute the modulus $r=\sqrt{a^{2}+b^{2}}$ and write it in exact form, simplifying the radical.
2 Write BOTH equations $\cos\theta=\frac{a}{r}$ and $\sin\theta=\frac{b}{r}$ , with their signs.
3 Locate the quadrant from these two signs, and SAY so explicitly on your paper.
4 Give $\theta$ as an exact value, stating the interval chosen if the question requires one.
5 Conclude with the full expression $z=r\,e^{i\theta}$ , and check by recomputing $r\cos\theta$ and $r\sin\theta$ .

Concluding sentence

$|z|=2$ , $\cos\theta=\frac{1}{2}$ and $\sin\theta=\frac{\sqrt{3}}{2}$ , so $\theta=\frac{\pi}{3}$ and $z=2\,e^{i\pi/3}$ .

The trap: Giving $\theta$ in degrees when the rest of the exercise is in radians, or the reverse. The exponential form requires RADIANS: $e^{i60}$ has nothing to do with $e^{i\pi/3}$ .

Marking: 0.5 mark for the modulus, 1 mark for the two equations and the quadrant, 0.5 mark for the final expression.

Solving an equation $z^{n}=a$

When to use it: The question asks for the $n$ th roots of a number, or to solve $z^{n}=a$ in $\mathbb{C}$ .

1 Put the right-hand side in exponential form, writing the argument WITH the term $2k\pi$ : $a=|a|\,e^{i(\alpha+2k\pi)}$ .
2 Set $z=\rho\,e^{i\varphi}$ and write the equation $\rho^{n}e^{in\varphi}=|a|e^{i(\alpha+2k\pi)}$ .
3 Identify the moduli, which gives $\rho=|a|^{1/n}$ , a single positive real value.
4 Identify the arguments, which gives $\varphi=\frac{\alpha+2k\pi}{n}$ , then list $k=0$ up to $n-1$ and no further.
5 Give the $n$ solutions, in exponential form and in algebraic form if the question asks, then check that their sum is zero.

Concluding sentence

The solutions of $z^{6}=64$ are $z_{k}=2\,e^{i\frac{2k\pi}{6}}$ for $k\in\{0, 1, 2, 3, 4, 5\}$ , that is six points forming a regular hexagon of radius $2$ .

The trap: Going beyond $k=n-1$ . For $k=n$ , you land back on $z_{0}$ : the roots repeat, and writing a seventh solution costs a rigour mark.

Marking: 0.5 mark for the $2k\pi$ , 0.5 mark for the modulus, 1 mark for the arguments, 1 mark for the complete list.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Recompute $a$ and $b$ from the exponential form

Once $r$ and $\theta$ are found, compute $r\cos\theta$ and $r\sin\theta$ : you must get exactly $a$ and $b$ back, SIGNS included. This is the check that catches every quadrant error.

For $-1-i$ , a $\theta=\frac{\pi}{4}$ would give $r\cos\theta=+1$ instead of $-1$ : the error is obvious.
Count the roots you found

An equation $z^{n}=a$ has exactly $n$ solutions, no more and no fewer. Counting before concluding is instant and catches both opposite errors, omission and repetition.

Four solutions announced for $z^{6}=64$ show that the $2k\pi$ was not pushed up to $k=5$ .
Check that the sum of the roots is zero

The $n$ $n$ th roots of a number always add up to $0$ , since they are the vertices of a regular polygon centred at the origin. A non-zero sum exposes a wrong root.

$2+(1+i\sqrt{3})+(-1+i\sqrt{3})+(-2)+(-1-i\sqrt{3})+(1-i\sqrt{3})=0$ : the six roots are right.
Check the modulus of a product or a power

$|z_{1}z_{2}|=|z_{1}||z_{2}|$ and $|z^{n}|=|z|^{n}$ . Comparing the modulus of the result with the one this rule predicts validates the calculation without redoing it.

$(1+i)^{8}$ must have modulus $\left(\sqrt{2}\right)^{8}=16$ : a result of modulus $2$ is wrong from the start.
Plot the point in the complex plane

Make a freehand sketch with both axes and the point. The quadrant can then be read without calculation, and the argument you found must agree with it.

An argument of $\frac{\pi}{4}$ announced for a point clearly at the bottom left is wrong: the sketch showed it before any calculation.

The typical problem, taken apart

From the algebraic form to the six sixth roots of 64

Let $z=1+i\sqrt{3}$ .

1) Write $z$ in exponential form.

2) Compute $z^{6}$ .

3) Solve the equation $w^{6}=64$ in $\mathbb{C}$ , and locate $z$ among the solutions.

The point

z=1+i\sqrt{3}

of the question, with real part

1

and imaginary part

\sqrt{3}

: question 1 asks for the length of the segment joining it to the origin and the angle of that segment with the real axis.

Step 1

$|z|=\sqrt{1^{2}+\left(\sqrt{3}\right)^{2}}=\sqrt{1+3}=2$ .

Why

The modulus always comes first: it normalizes the cosine and the sine, and it gives the radius of the circle of roots directly.

Step 2

$\cos\theta=\dfrac{1}{2}$ and $\sin\theta=\dfrac{\sqrt{3}}{2}$ : both are positive, so $z$ is in the first quadrant and $\theta=\dfrac{\pi}{3}$ .

Why

BOTH equations, not just one. The ratio $\frac{b}{a}=\sqrt{3}$ would also match $\frac{4\pi}{3}$ , in the third quadrant: the signs decide.

Step 3

$z=2\,e^{i\pi/3}$ , which can also be written $2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)$ .

Why

Immediate check: $2\cos\frac{\pi}{3}=1$ and $2\sin\frac{\pi}{3}=\sqrt{3}$ , we get $a$ and $b$ back. Ten seconds that validate the quadrant.

Step 4

De Moivre's theorem: $z^{6}=2^{6}\,e^{i\times 6\times\frac{\pi}{3}}=64\,e^{2i\pi}=64$ .

Why

The modulus is raised to the power, the argument is MULTIPLIED. Expanding $(1+i\sqrt{3})^{6}$ in algebraic form would take half a page and give the same $64$ .

Step 5

For the equation $w^{6}=64$ , write the right-hand side with the periodic term: $64=64\,e^{i(0+2k\pi)}$ .

Why

This is the decisive step of question 3. Without the $2k\pi$ , you find only $w=2$ and lose five solutions out of six.

Step 6

Setting $w=\rho e^{i\varphi}$ : $\rho^{6}=64$ gives $\rho=2$ , and $6\varphi=2k\pi$ gives $\varphi=\dfrac{k\pi}{3}$ .

Why

Identify the moduli, which are positive reals, separately from the arguments, which are defined modulo $2\pi$ . It is this second identification that produces the six angles.

Step 7

For $k=0$ to $5$ : $2$ , $1+i\sqrt{3}$ , $-1+i\sqrt{3}$ , $-2$ , $-1-i\sqrt{3}$ , $1-i\sqrt{3}$ . Beyond $k=5$ , the same ones come back.

Why

Six values of $k$ , six roots: counting them is the fastest check. For $k=6$ , the angle is $2\pi$ , that is $z_{0}$ again.

Step 8

Check: the sum of the six roots is $0$ , and $z=1+i\sqrt{3}$ is indeed the root for $k=1$ .

Why

The zero sum is guaranteed by the symmetry of the hexagon, and it tests all six values at once. Finding $z$ in the list closes the loop with question 2.

The conclusion, written out

$z=2\,e^{i\pi/3}$ , $z^{6}=64$ , and the six solutions of $w^{6}=64$ are $2$ , $1+i\sqrt{3}$ , $-1+i\sqrt{3}$ , $-2$ , $-1-i\sqrt{3}$ and $1-i\sqrt{3}$ , of which $z$ is the one for $k=1$ .

The classic mistake on this problem: Answering $w=2$ alone in question 3, because $64$ was written $64\,e^{i0}$ without the term $2k\pi$ . The answer is correct, but it is worth only one sixth of the marks.

Learn by heart

• Sum: ALGEBRAIC form. Product, quotient, power: EXPONENTIAL form.
• $z\bar{z}=|z|^{2}$ , a positive real: this is what makes a denominator real.
• The argument is fixed by the PAIR $\cos\theta=\frac{a}{r}$ and $\sin\theta=\frac{b}{r}$ , never by the tangent alone.
• $|z_{1}z_{2}|=|z_{1}||z_{2}|$ , but $|z_{1}+z_{2}|\leq|z_{1}|+|z_{2}|$ .
• De Moivre: $\left(re^{i\theta}\right)^{n}=r^{n}e^{in\theta}$ , modulus to the power, argument MULTIPLIED.
• $z^{n}=a$ has exactly $n$ solutions: write $a=|a|e^{i(\alpha+2k\pi)}$ BEFORE taking the root.
• The $n$ roots form a regular polygon of radius $|a|^{1/n}$ , and their sum is $0$ .
• $\sqrt{a}\sqrt{b}=\sqrt{ab}$ is FALSE if $a$ or $b$ is negative: write $2i$ , never $\sqrt{-4}$ .
• Conjugate roots: only if the coefficients of the polynomial are REAL.
• $|z-z_{0}|=R$ is the circle with centre $z_{0}$ and radius $R$ .

Frequently asked questions

How do you find the argument of a complex number?

First compute the modulus, then write two equations: the cosine equals the real part divided by the modulus, and the sine equals the imaginary part divided by the modulus. It is the signs of these two quantities together that fix the quadrant. The tangent alone is never enough, because it cannot tell a number from its opposite.

Why does the equation z to the power n equals a have n solutions?

Because the argument of a complex number is only defined up to a multiple of two pi. By writing the right-hand side with this periodic term before taking the root, you get n different angles, all within one full turn. The solutions are then spread evenly around a circle, at the vertices of a regular polygon.

How do you divide two complex numbers?

Multiply the numerator and the denominator by the conjugate of the denominator. The denominator then becomes the square of the modulus, a real number, and the quotient can be written in algebraic form. In exponential form it is even simpler: the moduli divide and the arguments subtract.

When should you use De Moivre's theorem?

Whenever you need to raise a complex number to a power greater than two, or to solve an equation where the unknown is raised to a power. Put the number in exponential form, raise the modulus to the power and multiply the argument by the exponent. Expanding in algebraic form would be much longer and much riskier.

Is the modulus of a sum the sum of the moduli?

No, except in a special case. The modulus behaves like a length: it distributes over a product, but not over a sum. There is only an inequality, the modulus of the sum being less than or equal to the sum of the moduli, with equality only when the two numbers have the same argument.

Can you write the square root of a negative number?

Better not. Over the complex numbers, every non-zero number has two square roots, so the usual notation loses its single meaning. Write the result directly with i, for example two i for minus four. The rule for multiplying two square roots, by the way, only holds for positive numbers.

Practise it

Corrected exercises: Complex numbers and De Moivre's theorem, 201-SN4-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

17 corrected exercises
170 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-sn4-complex-numbers. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: complex numbers and De Moivre's theorem (201-SN4-RE)

The essentials

Three forms, and what each one is for

The argument is fixed by a PAIR, never by a single ratio

De Moivre's theorem, the roots and the regular polygon

The mistakes that cost marks

1. Taking the arctangent without looking at the quadrant

2. Dividing without using the conjugate

3. Distributing the modulus over a sum

4. Applying De Moivre's theorem to the algebraic form

5. Giving only one nth root

6. Multiplying two square roots of negative numbers

7. Invoking conjugate roots for a polynomial with complex coefficients

8. Confusing zzˉz\bar{z}zzˉ with the modulus

Which method to choose

Which form to choose, depending on the operation

How to fix the argument, quadrant by quadrant

How the answer is expected to be written

Putting a complex number in exponential form

Solving an equation zn=az^{n}=azn=a

Check before you hand in

The typical problem, taken apart

From the algebraic form to the six sixth roots of 64

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Frequently asked questions

How do you find the argument of a complex number?

Why does the equation z to the power n equals a have n solutions?

How do you divide two complex numbers?

When should you use De Moivre's theorem?

Is the modulus of a sum the sum of the moduli?

Can you write the square root of a negative number?

Corrected exercises: Complex numbers and De Moivre's theorem, 201-SN4-RE

See also

Looking for a 201-SN4-RE tutor in Montreal?

8. Confusing $z\bar{z}$ with the modulus

Solving an equation $z^{n}=a$