1. Week 1: Complex Numbers, Alone and in Packs
Table of Contents | 2. Week 2: More on \(\exp\) and \(\log\) ≫Let’s recall the two basic ways we can express complex numbers: in rectangular form \(z=x+iy\), or in polar form \(z=re ^{i \theta }\). The quantities \(x, y, r, \theta \) all have special names:
- \(x\) is the real part of \(z\), denoted \(x = \Re z = \operatorname{Re} z\).
- \(y\) is the imaginary part of \(z\), denoted \(y= \Im z = \operatorname{Im} z\).
- \(r\) is the modulus of \(z\), denoted \(r = \left\lvert z \right\rvert\).
- \(\theta\) is “the” argument of \(z\), denoted \(\operatorname{arg}z\).
Of course, these all have graphical meanings: \(x\) and \(y\) correspond to the usual \(x\) and \(y\) coordinates we are all hopefully familiar with; \(r\) is the radial distance from \(z\) to the origin; and \(\theta \) is the angle swept out between \(z\) and the positive \(x\)-axis. Draw a picture! \(x\), \(y\), and \(z\) should make the sides of a right triangle.
A note about \( \theta \) — Euler is to blame for the famous equation \[e ^{i \theta } = \cos \theta + i \sin \theta .\] This formula, along with our graphical interpretation, indicate that \(\theta \) is not necessarily unique! Indeed, when \(z\) is nonzero, its argument is really an infinite set of real numbers, all \(2 \pi \) spaced apart from each other. This…gives us a lot of headaches, as we’ll see shortly.
Some computations will involve gross powers or products of complex numbers. In these cases, you should not sweat about spamming the binomial theorem or FOILing until your fingers bleed. Rather, you should convert things into polar coordinates, then multiply.
Problem 1.
Evaluate the following expressions.
- \((1+i)^8\).
- \(\left( \frac{\sqrt 3}{2} + \frac{i}{2} \right) ^{187}\).
- \(\left( -3i \right) ^{\frac{1}{5}}\) (you should get 5 answers!).
- \(\left( 1+\sqrt{-3} \right) ^{ \frac{1}{\pi }}\).
More on the Modulus
Let’s circle back to the modulus of \(z\), written \(\left\lvert z \right\rvert\) or as \(r\). From our favourite theorem (the Pythagorean theorem), we know that \(\left\lvert z \right\rvert^2 = x^2+y^2\); however, by using the difference of squares or a bit of divine foresight, one can also arrive at the expression \[\left\lvert z \right\rvert^2 = \left( x+iy \right) \left( x-iy \right) = z \cdot \overline{z}.\] This last quantity, \(\overline{z}\), is the complex conjugate of \(z\); one can envision it as a reflection of \(z\) over the horizontal axis. It is a very important quantity!!
There are several important relationships to point out, and you can prove these for fun if you want to:
Proposition 2. Helpful Complex Conjugation Identities
- \(\Re z = \frac{1}{2} \left( z+ \overline{z} \right)\).
- \(\Im z = \frac{1}{2} \left( z - \overline{z} \right)\).
- \( \overline{r e ^{i \theta }} = r e ^{- i \theta }\).
- \(\overline{z+w}= \overline{z} + \overline{w}\) .
- \(\overline{z \cdot w} = \overline{z} \cdot \overline{w}\).
- \(\overline{ \overline{z}} = z\).
- \(\left\lvert \overline{z} \right\rvert = \left\lvert z \right\rvert\).
- If \(\left\lvert z \right\rvert=1\), then \(\frac{1}{z}= \overline{z}\).
Try to interpret all of these statements geometrically, and draw a picture for each one! Better yet, provide proofs of all of these properties. Try to keep these in your back pocket — a lot of identities and equations rely on some subset of these relationships.
This alternative formula for the (squared) modulus can be very helpful in proving elementary properties of geometry, such as
Proposition 3. Triangle Inequality
\(\left\lvert z+w \right\rvert \leq \left\lvert z \right\rvert + \left\lvert w \right\rvert\) for all complex numbers \(z, w\).
Instead of sweating about \(\left\lvert z+w \right\rvert\), we should think about \(\left\lvert z+w \right\rvert^2\), which is easier to compute in terms of \(z\) and \(w\). Using the complex conjugate, we get \[\begin{align*} \left\lvert z+w \right\rvert^2 &= (z+w) \left( \overline{z+w} \right) \\ &= \left( z+w \right) \left( \overline{z}+ \overline{w} \right) \\ &= z \overline{z} + z \overline{w} + \overline{z} w + w \overline{w} \\ &= \left\lvert z \right\rvert^2 + z \overline{w} + \overline{z \overline{w}} + \left\lvert w \right\rvert^2 \\ &= \left\lvert z \right\rvert^2 + 2 \Re \left( z \overline{w} \right) + \left\lvert w \right\rvert^2. \end{align*}\] The cincher here is that \(\Re \left( z \overline{w} \right) \leq \left\lvert z \overline{w} \right\rvert = \left\lvert z \right\rvert \cdot \left\lvert w \right\rvert\)! Thus, we get that \[\left\lvert z+w \right\rvert^2 \leq \left\lvert z \right\rvert^2 + 2 \left\lvert z \right\rvert \left\lvert w \right\rvert + \left\lvert w \right\rvert^2 = \left( \left\lvert z \right\rvert+ \left\lvert w \right\rvert \right)^2.\] Taking squareroots yields the claim. \(\square\)
Exercise 4.
Prove that \(\left\lvert z+w \right\rvert = \left\lvert z \right\rvert+ \left\lvert w \right\rvert\) if and only if \(z = \lambda w\) for some real number \(\lambda \).
Hint
Exercise 5.
Fix \(a\in \mathbb{C}\). Show that \[\frac{\left\lvert z-a \right\rvert}{ \left\lvert 1- \overline{a} z \right\rvert} = 1\] if \(\left\lvert z \right\rvert = 1\) and \(\left\lvert 1- \overline{a}z \right\rvert\neq 0\).