Question

One of the values of ${{i}^{i}}$ is $\left( i=\sqrt{-1} \right)$
(a) ${{e}^{-\dfrac{\pi }{2}}}$
(b) ${{e}^{\dfrac{\pi }{2}}}$
(c) ${{e}^{\pi }}$
(d) ${{e}^{-\pi }}$

Answer 1

Hint: We first convert the given complex number as euler form then use the exponent property to solve further. Use definition of i and Euler’s formula as
${{e}^{ix}}=\cos x+i\sin x$ .

Complete step-by-step solution -
Definition of i:
i is an imaginary number which is solution of an equation:
${{x}^{2}}=-1\Rightarrow {{x}^{2}}+1=0$
Use Euler’s formula: ${{e}^{ix}}=\cos x+i\sin x$
The left-hand side can be written as $\text{cis}x$
So, $\text{cis}x=\cos x+i\sin x$
Let $x=\dfrac{\pi }{2}$
By substituting above $x$ value into expression, we get:
$\begin{aligned} & \text{cis}\dfrac{\pi }{2}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \\\ & \text{cis}\dfrac{\pi }{2}=i \\\ \end{aligned}$
We need value of ${{i}^{i}}$
So, the required expression can be written as:
${{\left( \text{cis}\dfrac{\pi }{2} \right)}^{i}}$
We know $\text{cis}\dfrac{\pi }{2}={{e}^{i\dfrac{\pi }{2}}}$
By substituting this into original equation, we get:
${{\left( {{e}^{i\dfrac{\pi }{2}}} \right)}^{i}}$
By using general algebraic identity:
${{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}$
By using above condition, we get:
${{e}^{{{i}^{2}}\dfrac{\pi }{2}}}$
We know i is solution of equation ${{x}^{2}}=-1$
So, ${{i}^{2}}=-1$
By substituting, we get:
${{e}^{-\dfrac{\pi }{2}}}$
So, ${{i}^{i}}={{e}^{-\dfrac{\pi }{2}}}$
Therefore ${{e}^{-\dfrac{\pi }{2}}}$ is value of required expression
Option (a) is correct.

Note: While using Euler’s formula be careful what to substitute in Cis and always keep $\sin x$ as imaginary if you keep $\cos x$ you may lead to wrong answer.
Idea of using Cis and again using back ${{e}^{ix}}$ is just for convenience you can directly use ${{e}^{ix}}$ .