Question

If $-\pi <\text{arg}\left(z\right)<-\frac{\pi }{2}$, then $\text{arg}\left(\overline{\mathrm{z}}\right)-\text{arg}\left(-\overline{\mathrm{z}}\right)$ is equal to

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

Answer 1

The correct option is A

$\pi$

Explanation for the correct option :

Suppose that $z=r\left(\cos \left(\theta \right)+i\sin \left(\theta \right)\right)$.

Then $\overline{z}$ which is the conjugate of $z$ will be

$\overline{z}=r\left(\cos \left(\theta \right)-i\sin \left(\theta \right)\right)$

Now, the $\text{arg}\left(\overline{z}\right)=\theta \dots \left(1\right)$

Similarly,

$-\overline{z}=r\left(i\sin \left(\theta \right)-\cos \left(\theta \right)\right)$

From this, we can conclude that,

$\text{arg}\left(-\overline{z}\right)=\theta -\pi \dots \left(2\right)$

From equation $\left(1\right)\text{and}\left(2\right)$, we have

$\begin{array}{rcl}\text{arg}\left(\overline{z}\right)-\text{arg}\left(-\overline{z}\right)& =& \theta -\left(\theta -\pi \right)\\ & =& \pi \end{array}$

Hence, the correct option is (A).