# 2nd degree equation (complex numbers)

1. Aug 22, 2007

### kasse

I just came across the eq.

z^2 - 2z + 1 - 2i

where z is a complex number. How do I solve this sort of eq.?

I tried to solve it as a normal 2nd degree eq., setting a=2, b=-2 and c=(1-2i), with z as the variable. This finally gave me the solutions

z(1) = -1 + sqrt(2i)

and

z(2) = -1 - sqrt(2i)

Can this be the correct solution? I had hoped for an answer involving i, not sqrt(i)...

2. Aug 22, 2007

### genneth

This is correct -- though you might want to find the solution to $$a+b i = \sqrt{i}$$ where a and b are real numbers. Remember that the point of the complex numbers to to be closed under things like taking square roots.

3. Aug 22, 2007

### Nesk

A relevant equation might be
$$\left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right)^2 = i$$

which can be found graphically by viewing the unit circle $$|z|=1$$ in the complex plane, and considering that if $$z = |z|_{\theta}$$, then $$z^n = |z|^n _{n \theta}$$, where $$\theta$$ is the angle between the lines corresponding to $$z$$ and $$1+0i$$.