Question

find the modulus and argument of the complex no. z = -1 -√3i

Answer 1

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\:z = - 1 - \sqrt{3}i$

To find the modulus and argument of complex number, we have to reduce to polar form.

So, Let assume that

$\rm :\longmapsto\: - 1 - \sqrt{3}i = r(cos\theta + isin\theta )$

$\rm :\longmapsto\: - 1 - \sqrt{3}i = rcos\theta + i \: rsin\theta$

On comparing, real and Imaginary parts, we get

$\rm :\longmapsto\:rcos\theta = - 1 - - - - (1)$

and

$\rm :\longmapsto\:rsin\theta = - \sqrt{3} - - - - (2)$

Squaring equation (1) and (2) and adding, we have

$\rm :\longmapsto\: {r}^{2} {cos}^{2}\theta + {r}^{2} {sin}^{2}\theta = 1 + 3$

$\rm :\longmapsto\: {r}^{2} ({cos}^{2}\theta + {sin}^{2}\theta) =4$

$\rm :\longmapsto\: {r}^{2} =4$

$\rm\implies \:r = 2 - - - (3)$

So, it means modulus of complex number is 2.

On substituting the value of r in equation (1) and (2), we get

$\rm :\longmapsto\:cos\theta = - \dfrac{1}{2}$

and

$\rm :\longmapsto\:sin\theta = - \dfrac{ \sqrt{3} }{2}$

$\rm :\longmapsto\:as \: sin\theta < 0 \: \: and \: \: cos\theta < 0$

$\rm\implies \:\theta \: \in \: {3}^{rd} \: quadrant$

$\rm\implies \:\theta = - \pi + \dfrac{\pi}{3}$

$\rm\implies \:\theta \: = \: - \: \dfrac{2\pi}{3}$

$\rm\implies \:arg(z) \: = \: - \: \dfrac{2\pi}{3}$

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More to Know

Argument Short Trick

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$