# Finding the roots of a polynomial with complex coefficients?

1. Jan 25, 2017

### Vitani11

1. The problem statement, all variables and given/known data
z2-(3+i)z+(2+i) = 0

2. Relevant equations

3. The attempt at a solution

Does the quadratic formula work in this case? Should you deal with the real and complex parts separately?

2. Jan 25, 2017

### Ray Vickson

Does algebra work with complex numbers? For complex numbers, do we have $a+b = b+a$, $a b = b a$, $a+(b+c) = (a+b)+c$, $a(bc) = (ab)c$, $a(b+c) = ab + ac$, and $a+0 = a$, $a 1 = a$? If so, then all the steps leading to the quadratic solution go through without change to the case of complex coefficients. In fact, in the derivation of the quadratic solution formula there was no mention of whether or not the coefficients were real.

Of course, when you need to express the final answer in the form $A + iB$ with real $A,B$ you might need to simplify something like
$$\frac{-(2+3i) \pm \sqrt{(2+3 i)^2 - 4 (5-2i)(7+6i) }}{2 (5-2i)}$$
and that will take some work. However, all the work before that is not changed by things being complex.

For practice, solve the example case $z^2 - (3+i) z + (2+i) = 0$ you started with.

3. Jan 25, 2017

### Vitani11

Great, thanks