Solving complex quadratic equations

You can solve quadratic equations over the complex numbers by completing the square¹ or using the quadratic formula.

Quadratic formula. For complex $a$, $b$ and $c$ (where $a\ne0$), the solutions to the equation $az^2+bz+c=0$ are $$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

In general, using the √ symbol is ambiguous because we don’t know which square root we are referring to, but here it is fine because we include both square roots.

Worked exercise: Find the solutions to the equation $z^2-4z+9=0$.
Solution: By completing the square, $$\begin{align*}
z^2-4z+4 &=-5 \\
(z-2)^2 &=-5. \end{align*}$$
Taking positive and negative square roots, we have $$\begin{align*}
z-2 &= \pm i \sqrt{5} \\
z &= 2+i\sqrt{5}
\end{align*}$$

Worked exercise: Find the solutions to the equation $z^2+(1-i)z+2+i=0$.
Solution: We will use the quadratic formula. The discriminant is equal to $$\bigtriangleup = b^2-4ac = (1-i)^2-4(2+i)=-8-6i.$$ We find that the square roots of $-8-6i$ are $\pm(1-3i)$. Hence, $$z= \frac{-(1-i)\pm(1-3i)}{2} = -i, -1+2i.$$

Problems

Find the solutions to the following equations.
1. $z^2+5=0$
2. $z^2+i=0$
3. $z^2-2z+2=0$
4. $z^2+6z+11=0$
5. $2z^2-8z+10=0$
Find the solutions to the following equations.
1. $3z^2+3z+8=0$
2. $(3+2i)z^2+(2\sqrt{2}+2\sqrt{2}i)z-2=0$
3. $iz^2+(-2-i)z+(6+24i)=0$

Recall that to complete the square on a quadratic equation $z^2+bz+c=0$, we move the constant to the other side, and add $\frac{b^2}{4}$ to get $\left(z+\frac{b}{2}\right)^2=\frac{b^2}{4}-c$.