Question

find the argument of...'i'​

Answer 1

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

Given complex number is

$\rm :\longmapsto\:i$

Let convert the given complex number to polar form.

So, Let assume that

$\rm :\longmapsto\:i = r(cosx + i \: sinx)$

where,

↝ r is the length of complex number

↝ x is argument of complex number

So, On comparing real and Imaginary parts, we get

$\red{\rm :\longmapsto\:rcosx = 0 \: - - - (1)}$

and

$\red{\rm :\longmapsto\:rsinx = 1 \: - - - (2)}$

On squaring equation (1) and (2) and add, we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2}x + {r}^{2} {sin}^{2}x = 0 + 1$

$\rm :\longmapsto\: {r}^{2}( {cos}^{2}x + {sin}^{2}x )= 1$

$\rm :\longmapsto\: {r}^{2}= 1$

$\rm \implies\:\boxed{ \tt{ \: r \: = \: 1 \: }}$

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

$\rm :\longmapsto\:cosx = 0$

and

$\rm :\longmapsto\:sinx = 1$

$\rm \implies\:\boxed{ \tt{ \: x \: = \: \frac{\pi}{2} \: }}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: argument \: of \: i \: = \: \frac{\pi}{2} \: }}$

More to know :-

Short Cut trick :-

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