We will look at the method of completing the square for solving equations of the form \[z^2 + px + q = 0\] where \(p\) and \(q\) are complex numbers.

The key step is to transform the equation \(z^2 + pz + q=0\) to the form \((z+r)^2 = s\) for some complex numbers \(r\) and \(s\). Note that the left-hand side is the square of \(z+r\). As \(s\) has two square roots, say \(s_1\) and \(s_2\), we must have \(z + r= s_1\) or \(z + r = s_2\), giving \(z = s_1 -r \) or \(z = s_2 -r \) as solutions.

The transformation can be obtained by noting that \begin{eqnarray} & & z^2 + pz + q =0 \\ & \Leftrightarrow & z^2 + pz = -q\\ & \Leftrightarrow & z^2 + pz + (p/2)^2 = - q + (p/2)^2\\ & \Leftrightarrow & (z + p/2)^2 = (p/2)^2 - q \end{eqnarray}

Example

Find all solutions to the equation \(z^2 + (-2+i)z -2i = 0.\)

Solution. Note that the equation is of the form \(z^2 + pz + q = 0\) with \(p = -2+i\) and \(q = -2i\).

Then \begin{eqnarray*} & & z^2 + (-2+i)z - 2i = 0 \\ & \Leftrightarrow & z^2 + (-2+i)z = 2i \\ & \Leftrightarrow & z^2 + (-2+i)z + \left ( \frac{-2+i}{2} \right)^2= \left ( \frac{-2+i}{2} \right)^2 + 2i \\ & \Leftrightarrow & \left(z + \left ( \frac{-2+i}{2} \right)\right)^2= \frac{3}{4} - i + 2i \\ & \Leftrightarrow & \left(z + \left ( -1 + \frac{1}{2}i \right)\right)^2= \frac{3}{4} + i \\ \end{eqnarray*}

Now, the square roots of \(\frac{3}{4} + i\) are \( 1 + \frac{1}{2} i\) and \(-1 - \frac{1}{2} i\). (See Exercise 1 for a way to obtain these.) Thus, the solutions are \( 1 + \frac{1}{2} i - \left ( -1 + \frac{1}{2}i \right)\) and \(-1 - \frac{1}{2} i - \left ( -1 + \frac{1}{2}i \right)\), or more simply, \(2\) and \(-i\).

Quick Quiz

Exercises

1. Let \(z = \frac{3}{4} + i\). The purpose of this exercise is to show that the square roots of \(z\) can be computed exactly.

   1. Write \(z\) in the form \(r \operatorname{cis}(\theta)\) where \(r\) is the modulus of \(z\) and \(\theta\) is the argument of \(z\).  

      \(\frac{5}{4} \operatorname{cis} \left(\tan^{-1} \frac{4}{3}\right)\)

   2. Recall that if \(0 \leq \theta \leq \pi\), then \[\cos\left(\frac{\theta}{2}\right)=\sqrt{ \frac{1+\cos(\theta)}{2}}\] and \[\sin\left(\frac{\theta}{2}\right)=\sqrt{ \frac{1-\cos(\theta)}{2}}.\] Use these formulas to convert the complex number \(r^\frac{1}{2}\operatorname{cis} \left(\frac{\theta}{2}\right)\) into rectangular form where \(r\) and \(\theta\) are from part (a).  

      Note that with \(\theta = \tan^{-1} \frac{4}{3}\), we have \(\cos \theta = \frac{3}{5}\) and \(\sin\theta = \frac{4}{5}.\) Thus, \(r^\frac{1}{2}\operatorname{cis} \left(\frac{\theta}{2}\right)\) is \( \frac{\sqrt{5}}{2} \sqrt{\frac{1+\frac{3}{5}}{2} } + \frac{\sqrt{5}}{2} \sqrt{\frac{1-\frac{3}{5}}{2} } i = 1 + \frac{1}{2} i.\)

   3. Give the square roots of \(z\).  

      \(1 + \frac{1}{2} i\) and \(-1 - \frac{1}{2} i\)