Question

If $z$ and $\omega$ are complex numbers such that $\left|z\omega \right|=1$, $\arg \left(z\right)-\arg \left(\omega \right)=\frac{3\mathrm{\pi }}{2}$ then find $\arg \left(\frac{1-2\overline{z}\omega }{1+3\overline{z}\omega }\right)$

• A. $\frac{\mathrm{\pi }}{4}$
• B. $-\frac{\mathrm{\pi }}{4}$
• C. $\frac{3\mathrm{\pi }}{4}$
• D. $-\frac{3\mathrm{\pi }}{4}$

Answer 1

The correct option is D

$-\frac{3\mathrm{\pi }}{4}$

Explanation for the correct option.

Step 1. Assume complex numbers $z$ and $\omega$.

It is given that $\left|z\omega \right|=1$, so let $\left|z\right|=r$ and $\left|\omega \right|=\frac{1}{r}$.

Also, it is given that $\arg \left(z\right)-\arg \left(\omega \right)=\frac{3\mathrm{\pi }}{2}$, so let $\text{arg}\left(z\right)=\theta$, then $\arg \left(\omega \right)=\theta -\frac{3\mathrm{\pi }}{2}$.

Now, the complex number $z$ and $\omega$ can be given as:

$z=r{e}^{i\theta };\omega =\frac{1}{r}{e}^{i\left(\theta -\frac{3\mathrm{\pi }}{2}\right)}$

Now the conjugate of $z=r{e}^{i\theta }$ is given as: $\overline{z}=r{e}^{-i\theta }$.

Step 2. Find $\overline{z}\omega$.

Multiply the complex number $\overline{z}$ and $\omega$.

$\begin{array}{rcl}\overline{z}\omega & =& r{e}^{-i\theta }\times \frac{1}{r}{e}^{i\left(\theta -\frac{3\mathrm{\pi }}{2}\right)}\\ & =& e^{-i\theta +i\theta -i\frac{3\mathrm{\pi }}{2}}\\ & =& e^{i\left(-\frac{3\mathrm{\pi }}{2}\right)}\\ & =& \cos \left(-\frac{3\mathrm{\pi }}{2}\right)+i\sin \left(-\frac{3\mathrm{\pi }}{2}\right)\left[e^{i\theta }=\cos \left(\theta \right)+i\sin \left(\theta \right)\right]\\ & =& 0+i\left(1\right)\\ & =& i\end{array}$

Step 3. Find the complex number $\frac{1-2\overline{z}\omega }{1+3\overline{z}\omega }$.

In the complex number $\frac{1-2\overline{z}\omega }{1+3\overline{z}\omega }$ substitute $i$ for $\overline{z}\omega$ and simplify.

$\begin{array}{rcl}\frac{1-2\overline{z}\omega }{1+3\overline{z}\omega }& =& \frac{1-2i}{1+3i}\mathbf{[}\overline{\mathbf{z}}\mathbf{\omega }\mathbf{=}\mathbf{i}\mathbf{]}\\ & =& \frac{\left(1-2i\right)\left(1-3i\right)}{\left(1+3i\right)\left(1-3i\right)}\\ & =& \frac{1-3i-2i+6i^2}{1-9i^2}\\ & =& \frac{1-5i+6\left(-1\right)}{1-9\left(-1\right)}\left[i^2=-1\right]\\ & =& \frac{1-5i-6}{1+9}\\ & =& \frac{-5-5i}{10}\\ & =& -\frac{1}{2}-\frac{1}{2}i\end{array}$

Step 4. Find the argument.

The complex number $\frac{1-2\overline{z}\omega }{1+3\overline{z}\omega }$i is the same as $-\frac{1}{2}-\frac{1}{2}i$.

The complex number $-\frac{1}{2}-\frac{1}{2}i$ is in third quadrant.

So, the argument is given as:

$\begin{array}{rcl}arg\left(-\frac{1}{2}-\frac{1}{2}i\right)& =& -\mathrm{\pi }+{\tan }^{-1}\left(\frac{-{\displaystyle \frac{1}{2}}}{-{\displaystyle \frac{1}{2}}}\right)\\ & =& -\mathrm{\pi }+{\tan }^{-1}\left(1\right)\\ & =& -\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\\ & =& -\frac{3\mathrm{\pi }}{4}\end{array}$

So, $\arg \left(\frac{1-2\overline{z}\omega }{1+3\overline{z}\omega }\right)=-\frac{3\mathrm{\pi }}{4}$.

Hence, the correct option is D.