Chapter 6.8, Problem 15E

To determine

Express the imaginary part of $\left(z\right)$ in terms of $z\text{ and }\overline{z}$ .

Answer to Problem 15E

The imaginary part of $\left(z\right)$ is $\frac{z-\overline{z}}{2}$ .

Explanation of Solution

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

Calculation: Consider that $z=x+iy$ where $x,y$ belongs to real numbers. And the real part of $\left(z\right)$ is $\left(x\right)$ , imaginary part is $\left(iy\right)$ .

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

Subtract (2) from (1),

  $\begin{array}{c}z-\overline{z}=\left(x+iy\right)-\left(x-iy\right)\\ z-\overline{z}=2iy\mathrm{....}\left(y\in R\right)\\ iy=\frac{z-\overline{z}}{2}\end{array}$

Hence the imaginary part of $\left(z\right)$ in terms of $z\text{ and }\overline{z}$ is $\frac{z-\overline{z}}{2}$ .

To determine

Show that $\left(z\right)$ is real if and only if $z=\overline{z}$ .

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. Now, according to question if $\left(z\right)$ is real then its imaginary part will be zero. Therefore,

  $\begin{array}{c}z=x+iy\mathrm{....}\left(1\right)\\ \text{such that, }\overline{z}=x-iy\mathrm{....}\left(\text{Conjugate}\right)\mathrm{....}\left(2\right)\\ z-\overline{z}=2iy\mathrm{....}\left(\begin{array}{l}\text{Subtract}\left(2\right)\\ \text{from }\left(1\right)\end{array}\right)\mathrm{....}\left(3\right)\end{array}$

Now from (3) $z-\overline{z}=2iy$ means if $\left(y=0\right)$ then,

  $\begin{array}{c}z-\overline{z}=0\\ z=\overline{z}\end{array}$

Hence proved that $z=\overline{z}$ if and only if imaginary part $\left(y\right)$ of $\left(z\right)$ equal to zero.