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I was working my way through a complex number worksheet with the promblem (3+2i)/z^2=1 and found a solution that is 1.817+0.550i but my other solution is -1.817-0.550i. These are not complex conjugates of each other which why I am confused, is this possible?

vince rabito
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    Only polynomials with real-valued coefficients are guaranteed to have conjugate roots, see https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem – Andreas Lenz Sep 22 '21 at 19:27
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    A polynomial with real coefficients has conjugate complex root pairs. This polynomial is $z^2-(3+2i)$ does not have real coefficients. – Thomas Andrews Sep 22 '21 at 19:27
  • The sum of the roots is $0$ by Vieta's relations, so if you found a root $z_0$ the other one is $z_1=-z_0,$. – dxiv Sep 22 '21 at 20:16

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If you have a polynomial equation with real valued coefficients then yes, for every complex root its complex conjugate is also a root.

However, the equation you are solving, after rewriting, becomes $$z^2 - (3+2i) = 0 $$ This polynomial does NOT have real valued coefficients. So there is no expectation that for every complex root its complex conjugate is also a root.

Lee Mosher
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