Chapter 6.8, Problem 13E

To determine

Show that $\left|\overline{z}\right|=\left|z\right|$ where $\left(z\right)$ is a complex number.

Explanation of Solution

Given information: The given complex number is $\left(z\right)$ .

Proof: Consider that $z=x+iy$ where $x,y$ belongs to real numbers.

  $\begin{array}{c}z=x+iy\mathrm{....}\left(\text{Assumption}\right)\mathrm{....}\left(1\right)\\ \text{such that, }\overline{z}=\overline{x+iy}\mathrm{....}\left(\text{Conjugate}\right)\\ \overline{z}=x-iy\mathrm{....}\left(2\right)\end{array}$

Evaluate $\overline{z}\text{ and }z$ respectively using (1) and (2),

  $\begin{array}{c}\left|\overline{z}\right|=\sqrt{{\left(x\right)}^2+{\left(-y\right)}^2}\\ =\sqrt{x^2+y^2}\mathrm{....}\left(3\right)\\ \left|z\right|=\sqrt{{\left(x\right)}^2+{\left(y\right)}^2}\\ =\sqrt{x^2+y^2}\mathrm{....}\left(4\right)\end{array}$

From (3) and (4) $\left|\overline{z}\right|=\left|z\right|$ .

Hence proved.