Question

i^i=
i=√-1
i(iota)
give ans in euler's form

Answer 1

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

Given expression is

$\rm :\longmapsto\: {i}^{i}$

Let first represent i in polar form

Let assume that,

$\rm :\longmapsto\:i = r(cosx + isinx) - - - - (1)$

can be rewritten as

$\rm :\longmapsto\:i = rcosx + i \: rsinx$

On comparing Real and Imaginary parts, we get

$\rm :\longmapsto\:r \: cosx = 0 - - - (2)$

and

$\rm :\longmapsto\:r \: sinx = 1 - - - (3)$

On squaring equation (2) and (3), and adding, we get

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

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

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

$\rm \implies\:\boxed{ \tt{ \: r \: = \: 1 \: }} - - - (4)$

On substituting the value of r, in equation (2) and (3), we get

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

and

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

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

On substituting the values of r and x in equation (1), we get

$\rm :\longmapsto\:i \: = \: 1\bigg[cos\dfrac{\pi}{2} + i \: sin\dfrac{\pi}{2} \bigg]$

$\bf\implies \:i = cos\dfrac{\pi}{2} + i \: sin\dfrac{\pi}{2}$

can be rewritten as

$\bf\implies \:i \: = \: \bigg[e\bigg]^{i\dfrac{\pi}{2}}$

Hence,

$\red{\rm :\longmapsto\: \: {i}^{i} \: \: }$

On substituting the value evaluated above, we get

$\bf\implies \: {i}^{i} \: = \: \bigg[e\bigg]^{i\dfrac{\pi}{2} \times i}$

$\bf\implies \: {i}^{i} \: = \: \bigg[e\bigg]^{ {i}^{2} \dfrac{\pi}{2}}$

$\bf\implies \: {i}^{i} \: = \: \bigg[e\bigg]^{ \: - \: \dfrac{\pi}{2}}$