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The problem statement is to solve the quadratic equation $$ z^2 + (\alpha + i\beta)z + \gamma + i\delta = 0. $$

I took the standard approach of $\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ but I am not not sure if this is the best way to approach the problem. I though about writing $z$ as $z = x + yi$, grouping, and completing the square either in reals/imaginary part or completing the square for $x$ and $y$. With the standard approach, I have $$ z = \frac{-\alpha - \beta i\pm\sqrt{\alpha^2 - \beta^2 + 2\alpha\beta i - 4\gamma - 4i\delta}}{2} $$ Now I could say $a + bi = \sqrt{\alpha^2 - \beta^2 + 2\alpha\beta i - 4\gamma - 4i\delta}$ and determine the complex number $a + bi$.

Is there a better way to do this problem? I feel like the above is too tedious. It is doable but I don't think it is optimal.

dustin
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    It seems you did not compute $(\alpha +i \beta)^2$ correctly. Other than that this is the common approach, IMO. Do you know how to calculate the sqrt of a complex number in algebraic form? – quid Dec 18 '14 at 23:50
  • The one you used $x + i y$ with real $x,y,$ incontrast to the trigonmetric form $r e^{i\phi}$ – quid Dec 18 '14 at 23:54
  • @quid I know how to do that. I just thought there would be a better approach to this problem or was at least hoping there would be. – dustin Dec 18 '14 at 23:55
  • Good. I would not know a general approach that it is better. But perhaps somebody else knows something. – quid Dec 18 '14 at 23:56

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You will not be able to get a simpler expression for $\sqrt{\alpha^2 - \beta^2 + 2 \alpha \beta i - 4 \gamma - 4 i \delta}$. Leave it in that form.

Robert Israel
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