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.

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- Aug 14th 2010, 03:15 AMGlitchFinding the roots of a complex number
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. - Aug 14th 2010, 03:39 AMVlasev
By the quadratic formula you should get

$\displaystyle \displaystyle z_{1,2} = \frac{3\pm(9-4(3-i))^{1/2}}{2} = \frac{3\pm(-3+4i)^{1/2}}{2} $

From here you need to use that the square root is multivalued. - Aug 14th 2010, 04:10 AMHallsofIvy
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 $\displaystyle \sqrt{r}e^{i\frac{\theta}{2}}= \sqrt{r}(cos(\frac{\theta}{2}+ \pi)+ i sin(\frac{\theta}{2}+ \pi))$ - Aug 14th 2010, 04:28 AMmr fantastic
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.

- Aug 14th 2010, 06:02 AMGlitch
- Aug 14th 2010, 01:49 PMVlasev
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.