# An Introduction to Complex Numbers

##### Age 16 to 18

Published 2001 Revised 2016

This is a short introduction to complex numbers written primarily for students aged from about 14 or 15 to 18 or 19. To understand the first few sections, it would be helpful to be familiar with polynomial equations (for example, solving $x^2 - 3x+2 = 0$), basic geometry (angles and lengths) and basic trigonometry (sine and cosine functions). For the later sections it would be helpful to be familiar with more advanced trigonometry (double angle formulae, for example), the exponential and logarithm functions ($e^x$ and $\log(x)$), vectors and matrices, and power series. The difficult bits are marked. (For students in the UK, the early sections should be accessible to anyone doing GCSE or above, but some of the later sections are A-level standard.) There are exercises throughout for you to test your understanding, with answers at the back.

I've tried to make the exercises less like the standard computational ones you get at school, but this means that some of them are quite hard.

Throughout this article, I've used both radians and degrees to measure angles. You might not have seen radians used before, they're just another way of measuring angles. Instead of measuring angles from $0^{\circ}$ to $360^{\circ}$ you measure them from 0 to $2\pi$. If you haven't seen this before, don't worry. Where possible I've included degrees and radians (although some bits only make sense in radians, you can skip these bits).

You can use wikipedia to look up any unfamiliar words or concepts.

Introduction

If you've done any quadratic equations, you'll know that there is a nice formula for the solution of the quadratic equation $a x^2 + b x + c = 0$, given by: $$x = \frac { -b\pm \sqrt{b^2 - 4a c}} {2a}$$

However, you'll also know that if $b^2 - 4ac$< 0 then there is no solution to the quadratic equation. Mathematicians like to have answers, so this annoyed them.

Imagine that the equation $x^2 + 5 = 0$ had a solution $\alpha$. If there was a solution (ignore the fact that there isn't a solution for the moment), we could work out all sorts of things in terms of $\alpha$ without actually knowing what it is.

For instance, what would $8\alpha^3 + 4\alpha^2 + 40\alpha + 20$ be? Well, $\alpha^2 + 5 = 0$, so $\alpha^2 = -5$. Multiplying this by $\alpha$, we get $\alpha^3 = -5\alpha$ . Putting this into the equation, we get:

\begin{eqnarray} 8\alpha^3 + 4\alpha^2 + 40\alpha + 20 &=& 8(-5\alpha) + 4(- 5)+40\alpha +20 \qquad &(\textrm{because} \quad \alpha^2 = -5 \quad \textrm{and} \quad \alpha^3 = -5 \alpha)& \\ &=& -40\alpha - 20 + 40\alpha + 20 \qquad &\textrm{(multiplying)}& \\ &=& 0 \qquad &\textrm{(simplifying)}& \end{eqnarray}

What happens if we substitute $x = 2\alpha$ into the polynomial $x^3 + x^{2} + 20x + 20$? Well, we get:
\begin{eqnarray} x^3 + x^2 + 20x + 20 &=& (2\alpha)^3 +(2\alpha)^2 + 20(2\alpha) + 20 &\textrm{(substituting)}& \\ &=& 8\alpha^3 + 4\alpha^2 + 40\alpha + 20 &\textrm{(expanding)}&\\ &=& 0 &\textrm{(using the equation above)}& \end{eqnarray}

So, what other equations can we solve armed with our new number $\alpha$?
Strangely enough, it turns out that if we are allowed to use $\alpha$, any polynomial equation has at least one root, but proving this is very difficult. If you do maths at university, this is one of the things you'll learn.

The Basics of Complex Numbers

If $i$ was a solution to the equation $x^2 + 1 = 0$ (again, ignore the fact that there is no solution) , then $i^2 + 1 = 0$, so $i^2 = -1$. If $\beta = i\sqrt{5}$, then $$\beta^2 = (i\sqrt{5})^2 = i^2 \times 5 = - 5 = \alpha^2$$

As you know, if $\alpha^2 = \beta^2$ then $\alpha = \pm \beta$. In other words, the number $\alpha$ above is just a multiple of $i$, $\alpha = \pm i \sqrt{5}$. So, a solution to the equation $x^3 + x^2 + 20x + 20 = 0$ is $x = 2\alpha = \pm 2 i \sqrt{5}$.

What we do now is pretend that the number $i$ really does exist (and why not?), but that it is very unlike any number we've seen before. This new number $i$ is called an imaginary number, because you have to imagine that it exists, and then you can do things with it. A complex number is a number $a+i b$, where $a$ and $b$ are the numbers you're familiar with (they're called real numbers). We can add two complex numbers to get a new complex number, $(a+i b)+(c+i d) = (a+c)+i(b+d)$. We can multiply them, $(a+i b)(c+i d) = a c+i b c+i a d+i^2 b d$ = $(a c-b d)+i(b c+a d)$.

Exercise 1 Work out $i^3 , i^4 , i^5 , i^6 , i^7$ and $i^8$. Do you see a pattern? What do you think $i^{443}$ is?

Exercise 2 Factorise the polynomial $x^2 - a^2$ (you don't need complex numbers for this)

Exercise 3 Factorise the polynomial $x^2 + a^2$ (you do need complex numbers for this)

Exercise 4 (easier) Work out $(1+i)^2$ and $(1+i)^3$.

Amazingly, we can also divide them, although it's not so obvious how to do that. What is $1/i$ for example? Well, we can multiply the top and bottom by the same number and get the same answer, because $a/i a$ = $1/i$ (the $a$'s cancel). Now set $a = i$ to get $1/i = i /i^2 = i /-1 = - i$. So $1 /i = - i$. Another way of seeing this is that $i^2 = - 1$, so if we divide both sides by $i$ we get $i = - 1 /i$ . What about $1 /(1+i)$ though? We can't multiply the top and bottom by $i$ now, because the bottom would still be a complex number which is no good. However, if we multiply the top and bottom by $1 - i$, we get: $$\frac{1}{1+i} = \frac{1- i}{(1+i)(1- i)} = \frac{1- i}{1+i- i+1} = \frac{1- i}{2} = \frac{1}{2} - i\frac{1}{2}$$

Excellent, but what about in general? What about $1/(a+i b)$? To answer this, we introduce something called the complex conjugate . We'll write the complex conjugate of $a+i b$ as $\textrm{Conj}(a+i b)$. We define it to mean $\textrm{Conj}(a+i b) = a - i b$. If $z = a+i b$, then $z\textrm{Conj}(z) = (a+i b)(\textrm{Conj}(a+i b)) = (a+i b)(a- i b) = a^2 + b^2$. But $a$ and $b$ are real numbers, so $z\textrm{Conj}(z)$ is a real number. Now we can answer the question of what is $[1/z]$ where $z$ is a complex number. $1 /z = \textrm{Conj}(z) /z\textrm{Conj}(z)$, where the top bit is a complex number and the bottom bit is a real number. Expanding it out we get $$\frac {1}{a + i b} = \frac {a -i b }{a^2 + b^2}$$

Exercise 5 Work out $$\frac {a^2 + b^2 }{a -i b}$$ If a complex number $z = a+i b$, we say that the real part of $z$, written $\textrm{Re}(z)$, is $a$, and the imaginary part of $z$, written $\textrm{Im}(z)$, is $b$. In other words, $\textrm{Re}(a+i b) = a$ and $\textrm{Im}(a+i b) = b$, if $a$ and $b$ are real numbers. Two complex numbers $z$ and $w$ are equal if $\textrm{Re}(z) = \textrm{Re}(w)$ and $\textrm{Im}(z) = \textrm{Im}(w)$, or in other words $a+i b = c+i d$ if $a = c$ and $b = d$.

Exercise 6 Express $\textrm{Re}(z^2)$ and $\textrm{Im}(z^2)$ in terms of $\textrm{Re}(z)$ and $\textrm{Im}(z)$.
Going back to our old friend the quadratic equation, if $a$, $b$ and $c$ are real numbers, the solution to $a x^2 + b x + c = 0$ is: $$x = \frac { -b\pm \sqrt{b^2 - 4a c}} {2a}$$ But now this formula works if $b^2 - 4ac < 0$ . Write $\Delta$ for $b^2 - 4ac$ (so $\Delta < 0$). $-\Delta > 0$ so we can work out $\sqrt{-\Delta}$ in the normal way. Also, $i^2 = -1$ so $i = \pm\sqrt{-1}$. Now $\sqrt{\Delta} = \sqrt{(-1)(-\Delta)} = \sqrt{-1}\sqrt{-\Delta} = \pm{i}\sqrt{(4a c - b^2)}$. So, if $b^2 - 4a c$ < 0, then $$x = \frac { -b\pm i \sqrt{4a c - b^2}} {2a}$$ Amazingly, the formula also works if $a$, $b$ and $c$ are complex numbers! However, we'll have to wait a bit to see what that means.

Exercise 7 (harder) If $w = 1 + i/\sqrt{2}$, work out $w^8$ [hint, work out $w^2$ first, then square that, then square that $((w^2)^2)^2 = w^8$]

Exercise 8 Find a solution to the equation $X^2 + 1 = 0$, using exercise 1.

Exercise 9 Find a solution to the equation $X^4 + 1 = 0$, using exercise 1. If you want a challenge, try finding all 4 solutions.

Exercise 10 (harder) Find a solution to the equation $X^2 + 2X + 2 = 0$.

Exercise 11 (very hard, for those who have done some trigonometry and used radians instead of degrees to measure angles only) Expand and simplify $(\cos(2\pi /2 )+ i\sin(2\pi /2 ))^2$. Expand and simplify $(\cos(2\pi /3 )+ i\sin(2\pi /3 ))^3$. Expand and simplify $(\cos(2\pi /4 )+ i\sin(2\pi /4 ))^4$. Can you guess what happens if you expand and simplify $(\cos(2\pi /n )+ i\sin(2\pi /n ))^n$? For those of you who have met mathematical induction , can you prove this?

The Argand Plane

We haven't yet got on to the most amazing thing about complex numbers, the geometric interpretation. As you know, a complex number $z$ can be written $a+i b$ where $a$ and $b$ are real numbers. If you've ever done vectors, this will look very familiar, a 2D vector can be written $a\mathbf{i} +b\mathbf{j}$ where $\mathbf{i}$ and $\mathbf{j}$ are the unit vectors. So, a complex number $z = a+i b$ corresponds to a point in a 2D plane, given by $a\mathbf{i} +b\mathbf{j}$. If you haven't done vectors using the notation above, $a+ib$ just corresponds to the point in 2D with $x$-coordinate $a$ and $y$-coordinate $b$. What about the sum of two complex numbers, $z+w$. It turns out that adding two complex numbers is the same as adding two vectors. If you don't know about how to add two vectors, look at the following picture: So, adding vectors corresponds to adding complex numbers. The 2D plane of complex numbers is called the Argand plane or Argand diagram. Well, that's nice, but not that great. The amazing thing is what happens when you multiply two vectors. Before we get on to that, we need a couple of new ideas.

The modulus $|z|$ of a complex number $z = a+i b$ is given by $|z|$ = $\sqrt{(a^2 + b^2)}$. By Pythagoras' theorem, this is just the length of the vector corresponding to $z$. There's another formula for $|z|$, using the complex conjugate we met earlier, namely $|z| = \sqrt{z\textrm{Conj}(z)}$ (just expand the right hand side out and see).

The argument $\arg(z)$ of a complex number $z$ is the angle between the vector corresponding to $z$ and the positive $x$-axis. If you've done some trigonometry, $\arg(z) = tan^{-1} (\textrm{Im}(z) /\textrm{Re}(z) )$ (unless $\textrm{Re}(z) = 0$). The following picture hopefully explains the modulus and argument of a complex number: Exercise 12 Let $w = (1+i)/\sqrt{2}.$ What are$|w|$ and $\arg(w)$?

Exercise 13 What is $|z|-|\textrm{Conj}(z)|$? Explain geometrically what $\textrm{Conj}(z)$ is in terms of $z$. Using this, what is $\arg(\textrm{Conj}(z))+\arg(z)$?

Now we can discuss the amazing thing about the geometry of complex numbers. If $z$ and $w$ are two complex numbers, then $|z w| = |z| |w|$ and $\arg(z w) = \arg(z)+\arg(w)$. In other words, if you multiply two complex numbers, you multiply their lengths and add their angles. We won't be able to prove the second equation until after the next section, but we can prove the first one. If $z = a+i b$ and $w = x+i y$, then $z w = (a x- b y)+i(a y+b x)$. So

\begin{eqnarray} |z w|^2 &=& (a x-b y)^2 +(a y+b x)^2 \\ &=& a^2x^2 -2a b x y+b^2y^2+a^2y^2 +2a b x y+b^2x^2 \\ &=& (a^2 + b^2)(x^2 + y^2) \\ &=& |z|^2 |w|^2 \end{eqnarray}

So $|z w| = |z| |w|$.

Exercise 14 What is $|z/w|$ in terms of $|z|$ and $|w|$? What is $\arg(z/w)$ in terms of $\arg(z)$ and $\arg(w)$? [Hint: $z = (z/w)(w)$]

Exercise 15 What is $|z^n|$ in terms of $|z|$? What is $\arg(z^n)$ in terms of $\arg(z)$? If you know about mathematical induction, prove your result.

Exercise 16 (for those who have done some trigonometry) If $z = r(\cos{\theta} +i\sin{\theta})$, what are $|z|$ and $\arg(z)$? Can you write $z$ in terms of $|z|$ and $\arg(z)$ for any complex number $z$? Can you use this to prove $\arg(z w) = \arg(z)+\arg(w)$?

Exercise 17 (very hard, for those who have done some trigonometry) What is $(\cos{\theta} +i\sin{\theta})^n$ in terms of $\cos{n\theta}$ and $\sin{n\theta}$?

Exercise 18 (very hard, for those who have completed the previous question only) Find a solution (not $z$ = 1) to the equation $z^n - 1 = 0$ in terms of $\cos(2\pi /n ) = \cos(360^{\circ}/n )$ and $\sin(2\pi /n ) = \sin(360^{\circ}/n )$. Find all the $n$ solutions in terms of this solution.

Polar Coordinates and De Moivre's Formula

There is another way of writing complex numbers apart from $a+i b$. Since a complex number is like a point in the complex plane, we can work out the distance of this point from the origin, $r$, and the angle that the line from the origin to the point makes with the $x$-axis, $\theta$. Once we've worked these out we can write a complex number as $(r,\theta)$, this is called the Polar Coordinate notation . These numbers $r$ and $\theta$ are just the modulus and argument of $x+i y$ that we met above.

So a complex number $z = a+i b$ can be written as $z = (|z|,\arg(z))$. Also, given a complex number $(r,\theta)$ we can change it into $x+i y$ notation (this is what Q1 is about) as $(r, \theta) = r(\cos\theta +i \sin\theta)$.

OK, so we can switch between two different notations for writing complex numbers, but what use is it? Well, it's very useful for a reason that will become apparent if you read the next section (which is a bit harder), but is also useful for a couple of other reasons.

I've already mentioned that $|z w| = |z| |w|$ and $\arg(z w) = \arg(z)+\arg(w)$. This makes it very easy to multiply and divide complex numbers written in polar coordinates, since $(r_1,\theta_1)(r_2, \theta_2) = (r_1 r_2 , \theta_1 + \theta_2)$.

Suppose $z = (r, \theta)$, what is $z^2$ in polar coordinates? Well, using the formula above, $z^2 = (r^2 ,2\theta)$. How about $z^3$, $z^4$, etc.? If we repeatedly use the above formula, we get that $z^3 = (r^3 ,3\theta)$, $z^4 = (r^4 ,4\theta)$. You can probably guess how the pattern continues, $z^n = (r^n ,n\theta)$. If $|z| = r = 1$ then $z = \cos\theta +i\sin\theta$, and so the formula in this case gives us: $$(\cos\theta +i\sin\theta)^n = \cos(n\theta) +i\sin(n\theta)$$

This is known as De Moivre's Formula and is the answer to exercise 17.

Exercise 19 Write $1+i$ in polar coordinates.

Exercise 20 Using the previous question, calculate $(1+i)^8$ in a couple of lines of working.

Exercise 21 (a bit tricky) Write $\cos(3\theta)$ in terms of $\cos(\theta)$. [Hint: use De Moivre's formula and the fact that $\cos^2\theta + \sin^2\theta = 1$]

Exponentials and Trigonometry (Advanced)

This section is much more difficult, you need to be able to understand measuring angles in radians instead of degrees. If you don't know about radians, skip to the next section.

Some functions have what is called a power series . If you've met these before, this section shouldn't be too difficult. If not, all you need to know is that for every number $\theta$ (an angle measured in radians) $$\cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \dots$$ and $$\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \dots$$

In fact, this is the reason that we use radians instead of degrees. You also need to know that the "exponential function'' $e^x$ (which is the number $e \approx 2.718281828\dots$ to the power of $x$) has a power series given by: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$$ What happens if we use this power series to calculate $e^{i\theta}$? So far, if you've used power series at all, you've probably only used them for real numbers, but they work just as well for complex numbers. Expanding $e^{i\theta}$ we get:

\begin{eqnarray} e^{i\theta} &=& 1 + (i\theta) + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \dots \\ &=& 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + \dots \\ &=& (1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} + \dots) + i(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} + \dots) \\ &=& \cos(\theta) + i\sin(\theta) \end{eqnarray}

The last line follows from looking at the power series for $\cos(\theta)$ and $\sin(\theta)$ above. So, we've proved that (for $\theta$ measured in radians), $e^{i\theta} = \cos\theta + i\sin\theta$ ! This is sometimes referred to as Euler's formula . You can use this to prove one of the most famous formulae in mathematics, which is $e^{i\pi} = - 1$, because $e^{i\pi} = \cos\pi +i\sin\pi = -1 + 0i = -1$.

So, what does this have to do with the polar coordinate notation? If we write $z = (r,\theta)$ then we can also write $z = {r}{e^{i\theta}}$ using the proof above.

Exercise 22 Prove that $|z w| = |z| |w|$ and $\arg(z w) = \arg(z)+\arg(w)$ using the ${r}{e^{i\theta}}$ formula.

Exercise 23 Find a solution to the equation $z^n = -1$ using the ${r}{e^{i\theta}}$ formula and $e^{i\pi} = - 1$.

Exercise 24 (a bit tricky) The hyperbolic cosine is defined to be $\cosh(z) = [e^z + e^{-z}] /2$. Find a power series expansion for $\cosh(z)$ and prove that $\cosh(i z) = \cos(z)$.

Cool stuff

Now we know the basics of how complex numbers work, what can we do with them?

Suppose we have a complex number $z = (r, \theta)$ in polar coordinates, and another complex number $w = (1, \phi)$ with modulus 1. The product of these two complex numbers is $z w = (r, \theta + \phi)$. In other words, $z w$ is $z$ rotated by an angle of $\phi$. We can use these to work out the matrix which rotates a vector by an angle $\phi$. Suppose $z = x+i y$ and $w = (1, \phi) = \cos\phi + i\sin\phi$, then $z w = (x+i y)(\cos\phi + i\sin\phi) = (x\cos\phi - y\sin\phi) + i(x\sin\phi + y\cos\phi)$. In other words, the $x$-coordinate (which is equivalent to the real part of a complex number) of a vector rotated by an angle $\phi$ is $x' = xcos\phi - y\sin\phi$ and the $y$-coordinate (the imaginary part) is $y' = xsin\phi - y\cos\phi$. We can represent this as a product of matrices:

$\left( \begin{array}{c} x'\\ y'\end{array}\right) = \left(\begin{array}{cc} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{array}\right) \left(\begin{array}{c} x \\ y \end{array}\right)$

In other words, the rotation matrix $M$ which rotates a vector $\textbf{v}$ to a vector $M\mathbf{v}$ by an angle $\phi$ is:

$\mathbf{M} = \left(\begin{array}{cc} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{array}\right)$

I mentioned at the beginning that you can solve any polynomial equation using complex numbers, this is known as the Fundamental Theorem of Algebra and a famous mathematician known as Gauss gave about 9 different proofs of this hundreds of years ago. More precisely, the fundamental theorem of algebra states that if $p(z) = z^n + a_{n-1}z^{n-1} + a_{n-2}z^{n-2} + \dots + a_{2}z^{2} + a_1z + a_0$ (where each of the $a_i$ is a complex number) then there is a complex number $w$ such that $p(w) = 0$. This is quite amazing really, since in creating complex numbers what we did was "invent'' a new number, a "solution'' to the equation $z^2 +1 = 0$, and we get the solutions to all polynomial equations for free!

Exercise 25 [hard] Use the fundamental theorem of algebra (stated above) and the Remainder Theorem (stated below) to prove, by induction, that a polynomial $p(z) = z^n + a_{n-1}z^{n-1} + a_{n-2}z^{n-2} + \dots + a_1z + a_0$ can be written $p(z) = (z - w_1)(z - w_2)\dots(z - w_n)$ for some complex numbers $w_i$. The degree of a polynomial is the highest power of $z$ in the polynomial, for example the degree of $z^3 + z^2 + z$ is 3 and the degree of $z^{18} + 2$ is 18. The Remainder Theorem says (or almost does) that if $p(w)$ = 0 then $p(z) = (z-w)q(z)$ for some polynomial $q(z)$ whose degree is one less than the degree of $p(z)$.

This section is quite difficult, and is for those who understood the section on exponentials and trigonometry.

If you know about the function $e^z$ and its power series, then you've probably also heard of the log function, which is its inverse. If not, the function $\log(z)$ is the unique (for real numbers) function such that $\log(e^z) = z$ and $e^{\log(z)} = z$.

Using this function, we can show that $x^y = e^{y\log(x)}$ for any positive real numbers $x$ and $y$. This is because (by the definition of log) $x = e^{\log(x)}$ , so $x^y = (e^{\log(x)})^y = e^{y\log(x)}$.

We can extend the idea of the log function to complex numbers, but unfortunately it is not unique (and there are other problems with it that I won't go into). For example, we know that $e^{2{\pi}{i}} = e^0 = 1$, so for complex numbers log(1) can be $2{\pi}{i}$ or 0! Since $x^y = e^{y\log(x)}$, this also means that $x^y$ is not unique when $x$ and $y$ are complex numbers.

However, there is a solution to this problem. We know that any complex number can be written $z = {r}{e^{i\theta}}$ , suppose that $\log(z) = x+i y$, then $e^{\log(z)} = z = {r}{e^{i\theta}} = e^{x+i y} = e^{x} e^{i y}$. So the modulus and argument of $e^{x} e^{i y}$ must be $r$ and $\theta$ respectively. In other words, $\log(z) = \log|z| +i\arg(z)$ where $\log|z|$ is unique because $|z|$ is a positive real number. I glossed over this problem earlier, but the argument of a complex number is not unique, the argument of $i$ could be $90^{\circ}$ or it could be $450^{\circ}$ , or in fact $(90+360n)^{\circ}$ for any integer $n$.

We say that the principal argument of $z$, written $\textrm{Arg}(z)$, is the unique angle $- \pi < \textrm{Arg}(z) \geq \pi$. Similarly, we can define the principal branch of the logarithm $\textrm{Log}(z) = \log|z| + i\textrm{Arg}(z)$. Using this, we can define the principal branch of complex exponentiation to be $x^y = e^{y\textrm{Log}(x)}$.

As an example, let's use this to work out the principal value of $i^i$ . We know that $e^{i\pi /2} = i$, and so $\textrm{Log}(i) = i\pi /2$ . So $i^i = e^{i\textrm{Log}(i)} = e^{i \times i\pi /2} = e^{-\pi /2}$, and this is a real number. In other words, we take one complex number to the power of another complex number and we get a real number!

Exercise 26 [very difficult indeed] Find all complex number solutions to the equation $x^y = y^x$ . [Hint: write $y = t x$ and see where you can go from there.]

The Basics of Complex Numbers

$i^3 = - i, i^4 = 1, i^5 = i, i^6 = - 1, i^7 = - i, i^8 = 1,$ etc. In general $i^{4n} = 1, i^{4n+1} = i, i^{4n+2} = - 1, i^{4n+3} = - i$, so $i^{443} = - i$.

Answer 2 $\quad (x- a)(x+a)$.

Answer 3 $\quad (x- i a)(x+i a)$.

Answer 4 $\quad (1+i)^2 = 2i, (1+i)^3 = 2i- 2$.

$$\frac {a^2 + b^2 }{a -i b}\ = \frac {(a^2 + b^2)(a + i b)}{a^2 + b^2}\ = a + i b$$

Answer 6 $\quad z = \textrm{Re}(z)+i\textrm{Im}(z)$, so $z^2 = \textrm{Re}(z)^2 - \textrm{Im}(z)^2 +2i(\textrm{Re}(z)\textrm{Im}(z))$. So $\textrm{Re}(z^2) = \textrm{Re}(z)^2 - \textrm{Im}(z)^2$ and $\textrm{Im}(z^2) = 2\textrm{Re}(z)\textrm{Im}(z).$

Answer 7 $\quad w^8 = 1$. This is called an eighth root of unity.

Answer 8 $\quad X = \pm i$.

Answer 9 $\quad X = \pm w = \pm 1+i /\sqrt{2}$. The other solutions are $- w$ , $i w$ and $- i w$.

Answer 10 The solution is, using the quadratic formula, $$x = \frac { -2\pm i \sqrt{4-8}} {2}\ = 1\pm \sqrt{-1}\ = -1 \pm i$$

Answer 11 Expanding each of the equations and simplifying gives 1 in each case (including the general case). Look at the section on De Moivre's formula for the proof.

The Argand Plane

Answer 12 $\quad |w| = 1$ and $\arg(w) = 45^{\circ} = \pi/4$.

Answer 13 $\quad|z|^2 = x^2 + y^2 = x^2 +(- y)^2 = |\textrm{Conj}(z)|^2$, so $|z| = |\textrm{Conj}(z)|$ and $|z| - |\textrm{Conj}(z)| = 0$. The complex conjugate is a reflection in the $x$-axis, so $\arg(\textrm{Conj}(z)) = - \arg(z)$ and hence $\arg(z)+\arg(\textrm{Conj}(z)) = 0$.

Answer 14 $\quad |z^n| = |z|^n$ , and we can prove this using induction. If $n = 1$ then it is obvious. If it is true for $n$, in other words we know that $|z^n| = |z|^n$ then $|z^{n+1}| = |z^{n} z| = |z^n| |z| = |z|^n |z| = |z|^{n+1}$, and so by induction it is true for all $n$. Similarly, $\arg(z^n) = n\arg(z)$.

Answer 15 Using the hint that $z = (z /w )w$ and the fact that $|u v| = |u| |v|$ we get that $|z| = |(z /w )| |w|$ , and so $|(z /w )| = |z| /|w|$ . Similarly, taking the argument of both sides, we get $\arg(z) = \arg(z /w )+\arg(w)$ and so $\arg(z /w ) = \arg(z)- \arg(w)$.

Answer 16 $\quad |z| = r$ and $\arg(z) = \theta$. So $z = |z| (\cos(\arg(z))+i\sin(\arg(z)))$. We can now prove that $|z w| = |z| |w|$ and $\arg(z w) = \arg(z)+\arg(w)$ by expanding out the brackets using this expansion of $z$ and $w$ and the trigonometric formula for $\cos(A+B) = \cos(A)\cos(B)- \sin(A)\sin(B)$ and $\sin(A+B) = \cos(A)\sin(B)+\sin(A)\sin(B)$.
So, if $|z| = r_1, |w| = r_2$, $\arg(z) = \theta_1$ and $\arg(w) = \theta_2$ then $z = r_1 (\cos\theta_1 +i\sin\theta_1)$ and $w = = r_2 (\cos\theta_2 +i\sin\theta_2)$. Expanding out

\begin{eqnarray} z w &=& r_{1} r_{2}\cos\theta_{1} +i\sin\theta_{1})(\cos\theta_{2} +i\sin\theta_{2})\\ &=& r_{1} r_{2} ((\cos\theta_{1} \cos\theta_{2} - \sin\theta_{1} \sin\theta_{2} )+i(\cos\theta_{1} \sin\theta_{2} + \cos\theta_{2} \sin\theta_{1} )) \\ &=& r_1 r_2 (\cos(\theta_{1} + \theta_{2} )+i\sin(\theta_{1} + \theta_{2} )) \end{eqnarray}

So $|z w| = r_1 r_2$ and $\arg(z w) = \theta_1 + \theta_2$.

Answer 17 There are various ways of answering this, you could calculate it directly using the formulas for $\cos(A+B)$ and $\sin(A+B)$. See the section on De Moivre's formula for an easier way.

Answer 18 The solution is $z = \cos(2\pi /n )+i\sin(2\pi /n) = cos(360^{\circ}/n)+i\sin(360^{\circ}/n)$ (which you can check works using the previous question). The other $n$ solutions are $z^2$, $z^3$,\dots, $z^n = 1$.

Polar Coordinates

Answer 19 $\quad 1+i = (\sqrt{2},45^{\circ}) = (\sqrt{2}, \pi /4 )$.

Answer 20 $\quad (1+i)^8 = (\sqrt{2},45^{\circ})^8 = (\sqrt{2^8} , 8 \times 45^{\circ}) = (16,360^{\circ}) = 16$.

Answer 21 Using De Moivre's formula, we have that $\cos(3\theta) + i\sin(3\theta) = (\cos\theta +i\sin\theta)^3$. We can expand out the right hand side to get $\cos^3\theta +i\cos^2\theta \sin\theta - \cos\theta \sin^2\theta - i\sin^3\theta$ which has real part $\cos^3\theta - \cos\theta \sin^2\theta$. We also know that $\sin^2\theta = 1 - \cos^2\theta$ , so the real part of $(\cos\theta +i\sin\theta)^3$ is $\cos^3\theta - \cos\theta (1 - \cos^2\theta) = 2\cos^3\theta - \cos\theta$. So $\cos(3\theta) = 2\cos^3\theta - \cos\theta$. You can use the same method to express $\cos(n\theta)$ or $\sin(n\theta)$ as a polynomial in $\cos\theta$ or $\sin\theta$.

Exponentials and Trigonometry (Advanced)

Answer 22 If $z = r_1 e^{i\theta_1}$ and $w = r_2 e^{i\theta_2}$ then $z w = r_1 r_2 e^{i\theta_1} e^{i\theta_2} = r_1 r_2 e^{i\theta_1 + \theta_2}$ and so $|z w| = |z| |w|$ and $\arg(z w) = \arg(z)+\arg(w)$.

Answer 23 We know $e^{i\pi} = - 1$ so $z^n = - 1$ means that $z^n = e^{i\pi}$. If $z = r e^{i\theta}$ then $z^n = r^n e^{{i}{n}\theta}$. So we need to solve $r^n e^{{i}{n}\theta} = e^{i\pi}$. Clearly $r = 1$ and $q = \pi /n$ work, so $z = e^{i\pi/n}$ is a solution.

Answer 24 By adding and expanding the series for $e^z$ and $e^{-z}$ we get that $$\cosh{z} = 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \dots$$ and so $$\cosh{i z} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} + \dots$$ which is just the power series for $\cos(z)$. You can use the same method to prove that if the hyperbolic cosine $\sinh(z) = [e^z - e^{-z}] /2$ then $\sinh(i z) = i\sin(z)$. Also, $\cos(i z) = \cosh(z)$ and $\sin(i z) = i\sinh(z)$.

Cool Stuff

Answer 25 We're trying to prove it by induction, so we have to first prove that it is true for $n = 1$. If $n = 1$ then $p(z) = z + a_0$ for some complex number $a_0$. In this case, $p(z)$ can be written $p(z) = (x- (- a_0 ))$ and so the theorem is true. Next we have to prove that if it is true for $n$ then it is true for $n+1$.

Suppose that $p(z)$ is a polynomial of degree $n+1$. By the fundamental theorem of algebra, there is a solution $w_{n+1}$. By the Remainder Theorem, $p(z) = (z- w_{n+1})q(z)$ where the degree of $q(z)$ is $n$. By induction, $q(z) = (z - w_1) \dots (z - w_n)$ and hence $p(z) = (z - w_1) \dots (z - w_n)(z - w_{n+1})$ and so the theorem is true by induction.

As a further exercise, you might like to try and prove the Remainder Theorem.

Answer 26 As I said, we start by writing $y = t x$, because if $x$ isn't zero (if $x = 0$ then $y = 0$ is the only solution, but $0^0$ is undefined so it isn't a solution) and $y$ is a solution then $y = y /x \times x$, or $t = y /x$. You might wonder why we introduce this spurious $t$, but just wait a minute and you'll see how useful it is.
Substituting $t x$ for $y$ in the equation $x^y = y^x$ we get $x^{t x} = (t x)^x$. Taking logs of both sides we get $t x\log(x) = x(\log(t)+\log(x))$. We know that $x \neq 0$ so we can divide both sides by $x$ to get $t\log(x) = \log(t)+\log(x)$. Rearranging, $\log(x)(t - 1) = \log(t)$, so $\log(x) = \log(t) / t- 1$. Taking exponentials, $x = e^{\log(t)/(t- 1)} = t^{1/(t- 1)}$. Also, $y = t x = t^{1/(t- 1)+1} = t^{t/(t - 1)}$. So we've found the solutions in terms of a parameter $t$.