Complex Functions: Domains, Image Sets and Inverses

I can imagine having to refer to these notes regularly, so I'm putting them here!

Image Sets

  1. State the domain A of f(z)=w.
  2. Rearrange so w is a function of z (to discover the condition under which w remains valid.

e.g., for f(z)=\frac{1}{z-1}:

    \begin{align*} f(A) =&\: \left\{ \frac{1}{z-1}\::\:z\in\mathbb{C}-\{1\}\right\} \\ f(A) =&\: \left\{w=\frac{1}{z-1}\::\:z\neq 1\right\} \\ f(A) =&\: \left\{w\::\: z=\frac{1}{w}+1\::\:z\neq 1\right\} \\ f(A) =&\: \{w\::\: w\neq 0\} \\ f(A) =&\: \mathbb{C}-\{0\} \\ \end{align*}

Domain of Combined Functions

Domain of combined functions are the intersection (A\cap B) of the domains of all component functions and that of the combined function. e.g:

f(z)=\frac{z-1}{z}\:,\:\: z\in\mathbb{Z}-\{0\}

g(z)=\frac{z}{z-1}\:,\:\: z\in\mathbb{Z}-\{1\}

\frac{f\left(z\right)}{g\left(z\right)} = \frac{z^{2}-2z+1}{z^{2}} \:,\:\: z\in\mathbb{Z}-\{0,\:1\}

Domain of Composite Functions

For f and g with domains A and B respectively, the domain of g\circ f is:

A-\{z\::\: f(z)\: \notin\: B\}

e.g., for:

f(z)=\frac{z-1}{z}\:,\:\: z\in\mathbb{C}-\{0\}

g(z)=\frac{z}{z-1}\:,\:\: z\in\mathbb{C}-\{1\}

\text{domain of }f\circ g = \text{domain of }g - \{z\::\:\frac{z}{z-1} \:\notin\: \mathbb{C}-\{0\}\}

\text{domain of }f\circ g = (\mathbb{C}-\{1\} ) - \{z\::\:\frac{z}{z-1} =0\}

\text{domain of }f\circ g = (\mathbb{C}-\{1\} ) - \{0\}

\text{domain of }f\circ g = (\mathbb{C}-\{0,\:1\} )


  1. Determine image set of f(z)=w.
  2. Invert f(z) to find a unique z in the domain of f.

For f(z)=\frac{1}{z-1}

f(A) = \{\frac{1}{z-1}\::\:z\in\mathbb{C}-\{1\}\}

f(A) = \{w=\frac{1}{z-1}\::\:z\:\neq\: 1\}

f(A) = \{w\::\: z=\frac{1}{w}+1\:\neq\: 1\}

f(A) = \{w\::\: w\:\neq\: 0\}

f(A) = \mathbb{C}-\{0\}

(all same as above for finding an image set)

z=\frac{1}{w}+1 gives a unique soluition in \mathbb{C}-\{0\}, hence f has a unique inverse rule: