How to solve these three complex-numbers equations

Question

I'm so stuck right now studying for my bachelor of science math exam. Please show (using i = imaginary unit):

1) i^(1/i) = e^(PI/2+k*2*PI)
2) (4*i)^(1/2) = { 2^1/2*(1+i) ; 2^1/2*(-1-i) }
3) i^(i*pi) = e^(-PI²/2)

I feel like I'm missing one thing for all of these. Huge thank you to anyone helping me out

Answer 1

We know that $i=e^{\frac{\pi}{2}i+k2\pi i}$.

For the first part, note that $$ \frac{1}{i} = \frac{i}{i^2}=-i $$ so we get $$ i^{\frac{1}{i}}=i^{-i}=e^{(i\frac{\pi}{2}+k2\pi i)\cdot -i}. $$ For the second part we want to get all the solutions of $$ x^2 = 4i. $$ For the third part we have $$ i^{i\pi} = e^{(i\frac{\pi}{2}+k2\pi i)\cdot \pi} $$ so you get infinitely many solutions there too.

Answer 2

Note that $$i^{\frac{1}{i}} = e^{\operatorname {Ln}{i^{\frac{1}{i}}}},$$ thus consider $$\operatorname {Ln}{i^{\frac{1}{i}}}=\frac{1}{i}\operatorname {Ln}{i}=\frac{1}{i}(\ln|i|+i \operatorname {Arg}{i}).$$ Since $$ |i|=1,\;\;\;\operatorname {Arg}{i}=\frac{\pi}{2}+2k\pi,$$ we have $$ \operatorname {Ln}{i^{\frac{1}{i}}}=\frac{\pi}{2}+2k\pi, \;\;\;k \in \mathbb{Z}, $$ therefore $$i^{\frac{1}{i}} =e^{\operatorname {Ln}{i^{\frac{1}{i}}}}=e^{\frac{\pi}{2}+2k\pi}.$$