Question

please help me in solving this question
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Answer 1

Answer:

8[cos pi/3+ (sin pi/3)i]

Answer 2

First we rationalize the denominator.

$\dfrac{-16}{1+\iota\sqrt3}\ =\ \dfrac{-16(1-\iota\sqrt3)}{(1+\iota\sqrt3)(1-\iota\sqrt3)}\ =\ \dfrac{-16(1-\iota\sqrt3)}{1+3}\ =\ 4(\iota\sqrt3-1)$

Now, let,

$x\cos\theta=-1\quad\quad;\quad\quad x\sin\theta=\sqrt3$

This gives  $\theta$  is in second quadrant.

$\dfrac{x\sin\theta}{x\cos\theta}=\dfrac{\sqrt3}{-1}\quad\implies\quad \tan\theta=-\sqrt3\\\\\\\therefore\ \theta=\dfrac{2\pi}{3}\quad\quad;\quad\quad x=2\\\\\\\therefore\ Z\ =\ 4(\iota\sqrt3-1)\ =\ 8\left(\dfrac{\sqrt3}{2}\iota-\dfrac{1}{2}\right)$

Hence Z in polar form is,

$\large\boxed{Z=8\left(\cos\left(\dfrac{2\pi}{3}\right)+\iota\sin\left(\dfrac{2\pi}{3}\right)\right)}$