# For z in the field of complex numbers

• October 25th 2012, 01:14 PM
patrickmanning
For z in the field of complex numbers
solve for z^2 = 2i
• October 25th 2012, 01:18 PM
richard1234
Re: For z in the field of complex numbers
Hint: $2i = 2(\cos {\frac{\pi}{2}} + i \sin {\frac{\pi}{2}}) = 2e^{i \frac{\pi}{2}}$.