zeros of a polynomial with complex coefficients

**Tala** · September 29th 2012, 01:12 PM

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

p(z) = z²-(1+2i)z + (- (3/2)+2i)

Is there a general method?

**johnsomeone** · September 29th 2012, 01:13 PM

yes - the quadratic equation.
Square roots of complex numbers are generally calculated by putting them into polar form first:

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

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

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

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

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