Question

FIND THE SQUARE ROOT OF i

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Answer 1

Answer:

$\sqrt{i}$

is the answer.

Step-by-step explanation:

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Answer 2

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

Let assume that

$\rm :\longmapsto\: \sqrt{i} = x + iy - - - (1)$

On squaring both sides, we get

$\rm :\longmapsto\:i = {(x + iy)}^{2}$

$\rm :\longmapsto\:i = {x}^{2} + {i}^{2} {y}^{2} + 2xyi$

$\rm :\longmapsto\:i = {x}^{2} - {y}^{2} + 2xyi$

So, on comparing real and Imaginary parts, we get

$\rm :\longmapsto\: {x}^{2} - {y}^{2} = 0 - - - - (2)$

and

$\rm :\longmapsto\:2xy = 1 - - - (3)$

We know, that

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{ {[ {x}^{2} - {y}^{2} ]}^{2} + {(2xy)}^{2}}$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{ {[0]}^{2} + {(1)}^{2}}$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{ {0 + 1}}$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{ {1}}$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 1 - - - - (4)$

On adding equation (2) and (4), we get

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

$\rm :\longmapsto\:{x}^{2} = \dfrac{1}{2}$

$\bf\implies \:x = \: \pm \: \dfrac{1}{ \sqrt{2} } - - - (5)$

On Subtracting equation (2) from equation (4), we get

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

$\rm :\longmapsto\:{y}^{2} = \dfrac{1}{2}$

$\bf\implies \:y= \: \pm \: \dfrac{1}{ \sqrt{2} } - - - (6)$

So, we have following options for the values of x and y

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{1}{ \sqrt{2} } & \sf \dfrac{1}{ \sqrt{2} } \\ \\ \sf - \dfrac{1}{ \sqrt{2} } & \sf - \dfrac{1}{ \sqrt{2} } \end{array}} \\ \end{gathered}$

So,

$\rm \implies\:\boxed{ \tt{ \: \sqrt{i} = \pm \: \bigg(\dfrac{1}{ \sqrt{2} } + i \: \dfrac{1}{ \sqrt{2} }\bigg) \: }}$

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Argument of complex number

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