# Thread: Finding the roots of a complex number

1. ## Finding the roots of a complex number

The question:
Find the roots (in Cartesian Form) of $z^2 - 3z + (3 - i) = 0$

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.

2. By the quadratic formula you should get

[LaTeX ERROR: Compile failed]

From here you need to use that the square root is multivalued.

3. It is simplest to take roots by using DeMoivre's formula: The nth roots of complex number $z= re^{i\theta}= r(cos(\theta)+ i sin(\theta))$, in polar form, are given by $\sqrt[n]{z}= r^{1/n}e^{i\frac{\theta+ 2k\pi}{n}}= r^{1/n}(cos(\frac{\theta+ 2k\pi}{n})+ i sin(\frac{\theta+ 2k\pi}{n}))$ where k= 0 to n-1 give the n roots. (r must be positive of course, and $r^{1/n}$ is the positive real n th root.)

In particular, if $a+ bi= r e^{i\theta}= r(cos(\theta)+ i sin(\theta))$ then the square root of a+ bi is given by $\sqrt{r}e^{i\frac{\theta}{2}}= \sqrt{r}(cos(\frac{\theta}{2})+ i sin(\frac{\theta}{2}))$ and $\sqrt{r}e^{i\frac{\theta}{2}}= \sqrt{r}(cos(\frac{\theta}{2}+ \pi)+ i sin(\frac{\theta}{2}+ \pi))$

4. Originally Posted by Glitch
The question:
Find the roots (in Cartesian Form) of $z^2 - 3z + (3 - i) = 0$

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.
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.

5. Originally Posted by Vlasev
By the quadratic formula you should get

[LaTeX ERROR: Compile failed]

From here you need to use that the square root is multivalued.
OK, I got this far. I'm not sure where I'm supposed to go from here. Is it a case of working out the square root using De Moivre's theorem, by converting to exponential form first?

6. Yea. You can do that. You will get two solutions (1+2i) and (-1-2i). Plugging both in the formula above, you will get a total of 2 different solutions.