You can solve quadratic equations over the complex numbers by completing the square^{1} 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.
- \(z^2+5=0\)
- \(z^2+i=0\)
- \(z^2-2z+2=0\)
- \(z^2+6z+11=0\)
- \(2z^2-8z+10=0\)

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