Question

If $z$ is a complex number of unit modulus and argument $\theta$, then $\arg \left(\frac{1+z}{1+\overline{z}}\right)$ is equal to

• A. $-\theta$
• B. $\frac{\mathrm{\pi }}{2}-\theta$
• C. $\theta$
• D. $\mathrm{\pi }-\theta$

Answer 1

The correct option is C

$\theta$

Explanation for the correct option.

Find the argument of the complex number.

It is given that for the complex number $z$, $\left|z\right|=1$ and $\arg \left(z\right)=\mathrm{\theta }$.

Now, for a complex number $z$, it is known that $z·\overline{z}={\left|z\right|}^2$. So it can be modified as

$$\begin{gathered} z·\overline{z}={\left|z\right|}^2 \\ \Rightarrow z·\overline{z}=1^2\left[\left|z\right|=1\right] \\ \Rightarrow z·\overline{z}=1 \\ \Rightarrow \overline{z}=\frac{1}{z}...\left(1\right) \end{gathered}$$

Now, in the complex number $\frac{1+z}{1+\overline{z}}$, use equation $1$ and substitute $\frac{1}{z}$ for $\overline{z}$.

$\begin{array}{rcl}\frac{1+z}{1+\overline{z}}& =& \frac{1+z}{1+{\displaystyle \frac{1}{z}}}\\ & =& \frac{1+z}{{\displaystyle \frac{z+1}{z}}}\\ & =& \frac{z\left(1+z\right)}{\left(1+z\right)}\\ & =& z\end{array}$

So the argument of the complex number is:

$\begin{array}{rcl}\arg \left(\frac{1+z}{1+\overline{z}}\right)& =& \arg \left(z\right)\end{array}$

And as $\arg \left(z\right)=\mathrm{\theta }$ so, $\arg \left(\frac{1+z}{1+\overline{z}}\right)=\theta$.

Hence, the correct option is C.