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

1. 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. $$2z^2-8z+10=0$$
2. 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. $$iz^2+(-2-i)z+(6+24i)=0$$

1. 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 $$\left(z+\frac{b}{2}\right)^2=\frac{b^2}{4}-c$$.