Question

Convert $\frac{\left(i+1\right)}{\cos \left({\displaystyle \frac{\pi }{4}}\right)-i\sin \left({\displaystyle \frac{\pi }{4}}\right)}$ in polar form

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

Answer 1

The correct option is C

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

Explanation for correct answer:

Step-1: Simplify the given data.

Given, $\frac{\left(i+1\right)}{\cos \left({\displaystyle \frac{\pi }{4}}\right)-i\sin \left({\displaystyle \frac{\pi }{4}}\right)}$$=$$\frac{\left(i+1\right)}{{\displaystyle \frac{1}{\sqrt{2}}}-i{\displaystyle \frac{1}{\sqrt{2}}}}$

$=$$\frac{\sqrt{2}\left(i+1\right)}{\left(1-i\right)}\times \frac{\left(1+i\right)}{\left(1+i\right)}$

$=$$\frac{\sqrt{2}{\left(i+1\right)}^2}{\left(1-i^2\right)}$

$=$$\frac{\sqrt{2}\left(i^2+1+2i\right)}{2}$

$=$$\sqrt{2}i$

Step-2: Finding $r$ and $\theta$ of the complex number.

Given, $z=0+\sqrt{2}i$

$\Rightarrow$ $r=\sqrt{0^2+{\left(\sqrt{2}\right)}^2}$

$\Rightarrow$ $r=\sqrt{2}$

and $\theta ={\tan }^{-1}\left(\frac{\sqrt{2}}{0}\right)$

$\Rightarrow$ $\theta =\frac{\pi }{2}$

Step-3: Finding polar form of the given complex number.

we know that the complex number in polar form.

$\Rightarrow$ $z=r\left(\cos \left(\theta \right)+i\sin \left(\theta \right)\right)$

$\Rightarrow$ $z=\sqrt{2}\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)$

Hence, correct answer is option $\left(C\right)$