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

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

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

z=\frac{1+7 i}{(2-i)^2}

, then the polar form of

z

View answer and explanation

Correct option: C

\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]

Option C (

\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]

) is correct because it represents the polar form of the given complex number after simplification.

Practice All Chapter MCQs →

{\left| {1 - i} \right|^x} = {2^x}

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}