Question

${i}^{i} \\ or \: \\ { (\sqrt{ - 1} )}^{ (\sqrt{ - 1}) }$
where i is iyota ​

Answer 1

Solution :-

Here we should know about the basic things which I will use to Solve this question.

I will use the concept of the complex no.

Now as we know every complex number is represented by Z.

and

$Z = r e^{(i) \theta}$

Where

r is distance from the origin to the point.

∅ is the angle from real axis

Co-ordinates :- ( real no. , n ( iota) )

Now our

$Z = i$

or

$Z = e^{(i) \frac{\pi}{2}}$

or

$i = e^{(i) \frac{\pi}{2}}$

Now coming back to our question :-

$\implies i^i = (e^{(i) \frac{\pi}{2}})^{i}$

$\implies i^i = e^{\left((i)\frac{\pi}{2}(i)\right)}$

$\implies i^i = e^{\left((i^2) \frac{\pi}{2}\right)}$

$\implies i^i = e^{\left((-1)\frac{\pi}{2}\right)}$

$\implies i^i = e^{ \left(\frac{-\pi}{2}\right)}$

Now value of

e = 2.71

π/2 = 1.5708

By solving :-

$\implies i^i = (2.71)^{(-1.5708)}$

$\implies i^i = \dfrac{1}{(2.71)^{(1.5708)}}$

$\implies i^i = \dfrac{1}{4.7875}$

$\implies i^i = 2.08873$

or

$\implies i^i \approx \dfrac{1}{5}$