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,
\[\begin{align}
  & a+ib=0+i \\
 & a+ib=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \\
\end{align}\]
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.
\[\begin{align}
  & \therefore {{i}^{i}}={{\left( 0+i \right)}^{i}}={{\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)}^{i}} \\
 & {{i}^{i}}={{\left( {{e}^{i\dfrac{\pi }{2}}} \right)}^{i}}={{e}^{i\times i\dfrac{\pi }{2}}}={{e}^{{{i}^{2}}\dfrac{\pi }{2}}} \\
\end{align}\]
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.