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: A complex number is of form $\left( a+ib \right)$. Thus for $i=0+i$. Now we use Euler’s identity, and get the value of $\theta$ in ${{e}^{i\theta }}$. Now raise this equation to the power of i and get the value of ${{i}^{i}}$.

Complete step-by-step answer:
In this equation, we need to find the value of ${{i}^{i}}$. We have been given that, $\left( i=\sqrt{-1} \right)$.
We know that a complex number is represented as $\left( a+ib \right)$, where a is the real part and b as the imaginary part. Hence, we can explain i as,

& a+ib=0+i \\\ & a+ib=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \\\ \end{aligned}$$ We can write like this because the value of $$\cos \dfrac{\pi }{2}=0$$ and that of $$\sin \dfrac{\pi }{2}=1$$. Now let us use Euler’s identity. In Euler’s identity the equality r is the Euler’s number, the base of natural logarithms, i is the imaginary unit, which by definition satisfies, $${{i}^{2}}=-1$$. Thus from Euler’s identity, we get $${{e}^{i\pi }}=-1$$ Hence by using Euler’s identity, $$\cos \theta +i\sin \theta ={{e}^{i\theta }}$$. Here, $$\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}={{e}^{i\dfrac{\pi }{2}}}$$ Hence, we got, $$\theta =\dfrac{\pi }{2}$$. $$\therefore i=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}={{e}^{i\dfrac{\pi }{2}}}$$ Now we can raise this equation to the power of i. $$\begin{aligned} & \therefore {{i}^{i}}={{\left( 0+i \right)}^{i}}={{\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)}^{i}} \\\ & {{i}^{i}}={{\left( {{e}^{i\dfrac{\pi }{2}}} \right)}^{i}}={{e}^{i\times i\dfrac{\pi }{2}}}={{e}^{{{i}^{2}}\dfrac{\pi }{2}}} \\\ \end{aligned}$$ We know that, $${{i}^{2}}=-1$$. Thus, $${{i}^{i}}={{e}^{\dfrac{-\pi }{2}}}$$. $$\therefore $$ Option (a) is the correct answer. Note: Here by Euler’s identity we got, $${{i}^{i}}={{e}^{\dfrac{-\pi }{2}}}$$, now this is a standard value which you can directly use as, $$i={{e}^{\theta }},\theta =\dfrac{\pi }{2}$$. Hence you can directly raise this to power I, without a lot of steps. Thus, $${{i}^{i}}={{e}^{\dfrac{-\pi }{2}}}$$ can be got in just a few steps.