solve for z^2 = 2i

Printable View

- Oct 25th 2012, 01:14 PMpatrickmanningFor z in the field of complex numbers
solve for z^2 = 2i

- Oct 25th 2012, 01:18 PMrichard1234Re: For z in the field of complex numbers
Hint: $\displaystyle 2i = 2(\cos {\frac{\pi}{2}} + i \sin {\frac{\pi}{2}}) = 2e^{i \frac{\pi}{2}}$.