Question

Show that the complex conjugation function $\mathit{f}\mathbf{:}\mathit{C}\mathbf{\to }\mathit{C}$(whose rule is$\mathit{f}\left(a+bi\right)\mathbf{=}\mathit{a}\mathbf{-}\mathit{b}\mathit{i}$ ) is a bijection.

Short answer

It is proved that$f$ is a bijection

Step-by-step solution

Definition

A function is said to be a bijection if it is injective and surjection both.

A function is injective if $f:B\to C$provided that whenever $f\left(a\right)=f\left(b\right)$in $C$, then $a=b$in $B$.

A function is said to be surjective, if $f:B\to C$provided that every element of C is the image under f of at least one element of B

Proof

Consider for every$z\in C$ , we have,

$$\begin{gathered} z=a+bi \\ =a-\left(-b\right)i \\ =f\left(a+\left(-b\right)i\right) \\ =f\left(a-bi\right) \end{gathered}$$

This implies that $f$is a surjection.

Apply injective definition as:

$$\begin{gathered} a-bi=c-di \\ a=cand-b=-d \\ a=candb=d \\ a+bi=c+di \end{gathered}$$

This implies that $f$is injection.

Hence, it is proved that $f$ is bijective.