# Complex Solutions

1. Jul 10, 2008

### lostinphys

Hi,
Could anyone please explain to me the procedure of obtaining complex solutions of equations where i is involved? For example z^3=i.
Many thanks!

2. Jul 10, 2008

### JohnSimpson

-i is one solution. I obtained that via more madness than method

3. Jul 10, 2008

### nicksauce

The usual thing to do would be to write the equal in exponential notation, ie $z = re^{i\theta}}$, $i=e^{i\frac{\pi}{2}}$, then for z^3 = i, we would have
$$r^3e^{3i\theta} = e^{i\frac{\pi}{2}}$$
From this you can conclude that r = 1. To find theta you take the logarithms,
$$3i\theta + 2ik\pi = i\frac{\pi}{2}$$
And then solve for theta, with different values of k (k as an integer).

4. Jul 10, 2008

### Santa1

You do it the same way you find the complex solutions of e.g. z^3=1

5. Jul 10, 2008

### lostinphys

thanks a bunch nicksauce :)