Finding the roots of a polynomial with complex coefficients?

[Vitani11]

Homework Statement

z²-(3+i)z+(2+i) = 0

Homework Equations

The Attempt at a Solution

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

 

[Ray Vickson]

Homework Statement

z²-(3+i)z+(2+i) = 0

Homework Equations

The Attempt at a Solution

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

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.

 

[Vitani11]

Great, thanks