The question:
Find the roots (in Cartesian Form) of $\displaystyle z^2 - 3z + (3 - i) = 0$
I'm not sure where to start. I tried applying the quadratic formula, but it doesn't seem to help. Any help would be great.
It is simplest to take roots by using DeMoivre's formula: The nth roots of complex number $\displaystyle z= re^{i\theta}= r(cos(\theta)+ i sin(\theta))$, in polar form, are given by $\displaystyle \sqrt[n]{z}= r^{1/n}e^{i\frac{\theta+ 2k\pi}{n}}= r^{1/n}(cos(\frac{\theta+ 2k\pi}{n})+ i sin(\frac{\theta+ 2k\pi}{n}))$ where k= 0 to n-1 give the n roots. (r must be positive of course, and $\displaystyle r^{1/n}$ is the positive real n th root.)
In particular, if [itex]a+ bi= r e^{i\theta}= r(cos(\theta)+ i sin(\theta))[/itex] then the square root of a+ bi is given by $\displaystyle \sqrt{r}e^{i\frac{\theta}{2}}= \sqrt{r}(cos(\frac{\theta}{2})+ i sin(\frac{\theta}{2}))$ and $\displaystyle \sqrt{r}e^{i\frac{\theta}{2}}= \sqrt{r}(cos(\frac{\theta}{2}+ \pi)+ i sin(\frac{\theta}{2}+ \pi))$
An alternative approach to the excellent approaches already suggested is to substitute z = a + ib, expand and equate the real and imaginary parts to zero. This gives two simultaneous equations to be solved for a and b, leading to the two solutions z = 2 + i and z = 1 - i.