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Polar form of $z= \frac{1+7i}{\left ( 2-i \right )^{2}}$ will be

Option 1)
$\sqrt{2}\left ( \cos \frac{\pi }{4} +i\sin \frac{\pi }{4}\right )$

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

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

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

Answers (1)

best_answer

$z=\frac{1+7i}{\left (2-i \right )^{2}}=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}=\frac{-25+25i}{25}$

$\Rightarrow \: z=-1+i$

$r=\left | z \right |=\sqrt{2}$ and $arg\left ( z \right )=\pi -\tan ^{-1}\left | \frac{1}{-1} \right |$

$\Rightarrow r=\sqrt{2}$ and $arg\left ( z \right )=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

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

$\therefore$ Option (B)

Polar Form of a Complex Number -

$z=r(cos\theta+isin\theta)$

- wherein

r= modulus of z and $\theta$ is the argument of z

Option 1)

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

This is incorrect

Option 2)

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

This is correct

Option 3)

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

This is incorrect

Option 4)

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

This is incorrect

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