MATH3401

Complex Analysis

Lecture 4

Roots of a complex number

For \(n>1\), \(n\)-th roots of \(z=re^{i\theta} \in \C_*\) means:
We want all \(w\in \C\) such that \(w^n=z\).

Notation: \(\exp(\xi)= e^{\xi}\).

Roots of a complex number

\(z\) has \(n\) distinct \(n\)-th roots:

\[{\small \begin{align*} & \left\{ r^{1/n}\exp\left(\frac{i\theta}{n}\right), r^{1/n}\exp\left(\frac{i\theta}{n} + \frac{i2\pi}{n}\right), \right. \\ & \quad \left. r^{1/n}\exp\left(\frac{i\theta}{n} + \frac{i4\pi}{n}\right), \ldots, r^{1/n}\exp\left(\frac{i\theta}{n} + \frac{i2(n-1)\pi}{n} \right) \right\} \end{align*}} \]

Functions and mappings

\(\Omega \subseteq \C\): A function \(f:\Omega \ra \C\) can be viewed as a mapping on \(\Omega\), the domain of \(f\).

If \(\Omega \) is not specified, take \(\Omega \) as large as possible.

For example for \(f(z)=\dfrac{1}{z}\), the domain is \(\C_*\) and \(f:\C_*\ra \C\).

We can write: \(f:z \mapsto \dfrac{1}{z}\); or \(w=\dfrac{1}{z}\); or just \(\dfrac{1}{z}\).

Functions and mappings

Usual notation: \(w: (x,y) \mapsto (u,v)\) i. e., \[ \begin{align*} w(x+iy) &= u(x+iy)+iv(x+iy)\\ & \text{ or} & \\ w(x, y) &= u(x,y)+iv(x, y) \end{align*} \]

Functions and mappings

Functions and mappings

\(*\, \Omega = \) domain of \( f = \) dom\(\left(\,f\right)\).

\(*\,\) Range\(\left(\,f\right)= f\left(\Omega\right)\) \[ = \left\{ w: w=f(z)\text{ for some }z\in\Omega\right\}. \]

\(*\, f^{-1}\left(\xi \right) = \left\{ z\in\C: f(z)=\xi \right\}\) inverse of \(f\) (not \(1/f\)).
NOT necessarily a function!

Functions and mappings

Example 1   -   \(f(z)=\dfrac{1}{z}\).

dom\(\left(\,f\right) = \C_*\),

\( \;f^{-1}(\xi)=\dfrac{1}{\xi}\) is a function.

Functions and mappings

Example 2   -   \(g(z)=\dfrac{1}{1-|z|^2}\).

dom\(\left(g\right) = \left\{ z:|z|\neq 1\right\}\),

\(g: \left\{ z:|z|\neq 1\right\}\ra \R\).

Inverse is not a function.

Check: If \(g(z)=\rho\), then \(g(e^{i\theta}z)=\rho\) for any \(\theta \in \R\). 📝

Functions and mappings

Example 3   -   \(h(z)=z^2\).

\(h:\C\ra \C\),

inverse is not a function.

Aim: Get a geometric picture

of a given function

E. g. \(w = 1 + z\) moves each point one (unit) to the right.

E. g. \(re^{i\theta} \mapsto re^{i\left(\theta + \pi/2\right)}\) rotates through an angle of \(\pi/2\) in the positive (counterclockwise) direction about 0.

Basic idea with new mappings

Break it down into compositions of known/easy maps.

Example 1: Linear transformation

\(w=Az+b\) with \(A,b\in \C, A\neq 0\).

Split up into:

  1. dilatation & rotation;
  2. translation.

Basic idea with new mappings

Example 1 (cont): Linear transformation

(i) \(z\mapsto Az\): Write \(A=ae^{i\alpha}\) with \(\alpha\in \R, a \in \R_+\).

So \(re^{i\theta} \mapsto \left(a r\right)e^{i\left(\theta + \alpha \right)}\) geometrically dilates (expands or contracts) modulus by a factor of \(a=|A|\), and rotates through an angle of \(\alpha=\arg \,A\).

Basic idea with new mappings

Example 1 (cont): Linear transformation

(ii) \(z\mapsto z + b\): Write \(b=b_1+b_2 i\) with \(b_1,b_2\in \R\).

Then translate \(b_1\) to the right and \(b_2\) up.

\(z \mapsto A z + b\): compose (i) and (ii).

Note: domain is \(\C\).

Next: \(\dfrac{1}{z}\).

Credits