Question

Represent the complex number (-1-√3i) in the polar form

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assumption

z = (-1 - √3i)

It is clear that (-1 - √3i) lies in 3 Quadrant

z = r(cosθ + isinθ)

Now,

rcosθ = -1

rsinθ = -√3

Now here,

$\textbf{\underline{Squarring\;both\;sides :-}}$

r² = 4

r = √4

r = 2

Hence,

${\boxed{\sf\:{cos\theta=\dfrac{-1}{2}}}}$

Also,

${\boxed{\sf\:{sin\theta=\dfrac{-\sqrt{3}}{2}}}}$

tanθ = √3

tanα = |tanθ| = √3

${\boxed{\sf\:{\alpha=\dfrac{\pi}{3}}}}$

Then,

$\textbf{\underline{It\;lies\;in\;3\;Quadrant}}$

Hence,

θ = (π - α)

${\boxed{\sf\:{\dfrac{\pi}{3}-\pi=\dfrac{-2\pi}{3}}}}$

Thus,

$\Large{\boxed{\sf\:{z=2[cos (\dfrac{-2 \pi}{3})+isin(\dfrac{-2 \pi}{3})}}}$

Answer 2

$\huge{\boxed{\mathbb{\blue{Polar\:Form}}}}$

$z = 2( \cos( \frac{2\pi}{3} - i \sin( \frac{2\pi}{3} ) )$

Here r = |z|

$|z| = \sqrt{ {re(z)}^{2} + {im(z)}^{2} }$

$|z| = \sqrt{ { - 1}^{2} + ( { - \sqrt{ 3} }^{2}) }$

$|z| = \sqrt{1 + 3}$

$|z| = 2$

Let alpha be the acute angle given by tan alpha

$\tan( \alpha ) = | \frac{im(z)}{re(z)} |$

$\tan( \alpha ) = | \frac{ - \sqrt{3} }{ - 1} |$

$\tan( \alpha ) = \sqrt{3}$

Clearly (-1,-√3 ) is in 3rd quadrant .

Therefore argument is given by

$\theta = - (\pi - \alpha )$

$\theta = - (\pi - \frac{\pi}{3} )$

$\theta = - \frac{2\pi}{3}$

Now polar form of compex number is

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

$z = 2( \cos( \frac{2\pi}{3} - i \sin( \frac{2\pi}{3} ) )$

hope it helps