Question

If $z=\frac{1+7i}{{\left(2-i\right)}^2}$ then the polar form of $z$ is:

• A. $\sqrt{2}\left(\cos \frac{3\mathrm{\pi }}{4}+i\sin \frac{3\mathrm{\pi }}{4}\right)$
• B. $\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{4}+i\sin \frac{\mathrm{\pi }}{4}\right)$
• C. $\sqrt{2}\left(\cos \frac{7\mathrm{\pi }}{4}+i\sin \frac{7\mathrm{\pi }}{4}\right)$
• D. $\sqrt{2}\left(\cos \frac{5\mathrm{\pi }}{4}+i\sin \frac{5\mathrm{\pi }}{4}\right)$

Answer 1

The correct option is A

$\sqrt{2}\left(\cos \frac{3\mathrm{\pi }}{4}+i\sin \frac{3\mathrm{\pi }}{4}\right)$

Explanation for the correct option.

Step 1. Write the complex number in standard form.

First, simplify the denominator of the complex number $z=\frac{1+7i}{{\left(2-i\right)}^2}$.

$\begin{array}{rcl}z& =& \frac{1+7i}{{\left(2-i\right)}^2}\\ & =& \frac{1+7i}{4-4i+i^2}\\ & =& \frac{1+7i}{4-4i+(-1)}\left[i^2=-1\right]\\ & =& \frac{1+7i}{3-4i}\end{array}$

Now, multiply the numerator and denominator by $3+4i$.

$\begin{array}{rcl}z& =& \frac{\left(1+7i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}\\ & =& \frac{3+4i+21i+28i^2}{9-16i^2}\\ & =& \frac{3+25i+28\left(-1\right)}{9-16\left(-1\right)}\left[i^2=-1\right]\\ & =& \frac{-25+25i}{25}\\ & =& -1+i\end{array}$

Step 2. Find the polar representation of the complex number.

Multiply the complex number $z=-1+i$ by $\frac{\sqrt{2}}{\sqrt{2}}$.

$\begin{array}{rcl}z& =& \frac{\sqrt{2}}{\sqrt{2}}\left(-1+i\right)\\ & =& \sqrt{2}\left(-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)\\ & =& \sqrt{2}\left(\cos \frac{3\mathrm{\pi }}{4}+i\sin \frac{3\mathrm{\pi }}{4}\right)\left[\cos \frac{3\mathrm{\pi }}{4}=-\frac{1}{\sqrt{2}},\sin \frac{3\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}\right]\end{array}$

So, the polar representation of $z=\frac{1+7i}{{\left(2-i\right)}^2}$ is $\sqrt{2}\left(\cos \frac{3\mathrm{\pi }}{4}+i\sin \frac{3\mathrm{\pi }}{4}\right)$.

Hence, the correct option is A.