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Solving quadratic equations with complex coefficients 3

Determine the solutions of the equation $$ z^2 - (1+3i)z - 2 + 2i = 0 $$
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To proceed, let us use the quadratic formula $$ \begin{align} z &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ &= \frac{(1+3i) \pm \sqrt{(1+3i)^2 - 4(-2+2i)}}{2}\\ &= \frac{(1+3i \pm \sqrt{-2i}}{2} \end{align} $$ Again, without appealing to modulus-argument methods, we can directly calculate $\sqrt{-2i}$.

Let $\sqrt{-2i} = a+ bi$ then $a^2- b^2 + 2abi = -2i$. This gives $a^2 = b^2$ and $2ab = -2 \Rightarrow b = \frac{-1}{a}$. Substitution of this yields $$ a^2 - \frac{-1}{a^2} = 0 \Rightarrow a^4 = 1 $$ Let $a = 1$, then $b = -1$, so one solution of the simultaneous equation is $a + bi = 1 -i$ similarly $a+bi = -1 + i$ is a solution.

It follows that the solutions to the original equation are therefore $$ \begin{align} z &= \frac{(1+3i \pm \sqrt{-2i}}{2} \\ &= \frac{1+3i \pm (1-i)}{2} \\ \end{align} $$ This gives $z = 2i$ or $z = 1+i$ as solutions.