Calculate {i}^{\frac{3}{4}}

I tried with

i=cos\frac{pi}{2}+isin\frac{\pi}{2}

i^3=cos\frac{3pi}{2}+isin\frac{3\pi}{2}

i^3=-i

\sqrt[4]{i^3}=\sqrt[4]{-i}

What should I do next?

It would be a lot easier if you started with i=e^(i.pi(2n+1/2)) (.= multiply, n is any integer)

Then i^3/4=e^(i.pi(3n/2 + 3/8)). You will have 4 distinct results for n=0,1,2,3.

Ok, thank you..

Modulus is one so DeMoivre's Theorem should yield ...
i^ [3/4] = cos [3 pi / 8] + i sin [3 pi / 8]