How do I find the roots of this polynomial with complex coefficients:

p(z) = z^{2}-(1+2i)z + (- (3/2)+2i)

Is there a general method?

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- Sep 29th 2012, 01:12 PMTalazeros of a polynomial with complex coefficients
How do I find the roots of this polynomial with complex coefficients:

p(z) = z^{2}-(1+2i)z + (- (3/2)+2i)

Is there a general method? - Sep 29th 2012, 01:13 PMjohnsomeoneRe: zeros of a polynomial with complex coefficients
yes - the quadratic equation.

Square roots of complex numbers are generally calculated by putting them into polar form first:

$\displaystyle x + iy = \sqrt{x^2 + y^2}e^{i\tan^{-1}(y/x)}$ when $\displaystyle x>0$,

$\displaystyle x + iy = \sqrt{x^2 + y^2}e^{i(\pi + \tan^{-1}(y/x))}$ when $\displaystyle x<0$,

$\displaystyle x + iy = \sqrt{x^2 + y^2}e^{i \pi/2 }$ when $\displaystyle x=0, y>0$,

$\displaystyle x + iy = \sqrt{x^2 + y^2}e^{- i \pi/2 }$ when $\displaystyle x=0, y<0$.