Chapter 6.8, Problem 12E

To determine

With the help of $\frac{1}{z}=\frac{\overline{z}}{{\left|z\right|}^2}$ evaluate the value of $\frac{1}{z}$ for $z=\frac{\sqrt{3}}{2}-\frac{1}{2}i$ .

Answer to Problem 12E

The value of $\frac{1}{z}$ is equal to $\frac{\sqrt{3}}{2}+\frac{1}{2}i$ .

Explanation of Solution

Given information: It is given that,

  $z=\frac{\sqrt{3}}{2}-\frac{1}{2}i\mathrm{....}\left(1\right)$

Formula used: The formula used to evaluate $\frac{1}{z}$ is written as,

  $\frac{1}{z}=\frac{\overline{z}}{{\left|z\right|}^2}\mathrm{....}\left(2\right)$

Calculation: Since $z=\frac{\sqrt{3}}{2}-\frac{1}{2}i$ therefore $\overline{z}=\frac{\sqrt{3}}{2}+\frac{1}{2}i$ . Now evaluate ${\left|z\right|}^2$ using (1). Therefore from (1),

  $\begin{array}{c}{\left|z\right|}^2={\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(-\frac{1}{2}\right)}^2\\ =\frac{3}{4}+\frac{1}{4}\\ =1\end{array}$

Now from (2) evaluate $\frac{1}{z}$ ,

  $\begin{array}{c}\frac{1}{z}=\frac{\overline{z}}{{\left|z\right|}^2}\\ =\frac{\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)}{\left(1\right)}\\ =\frac{\sqrt{3}}{2}+\frac{1}{2}i\end{array}$