Gary Liang Notes

Free online notes for Year 11 and 12 HSC, VCE and QCE students

Square roots of complex numbers

Square roots of a negative real number

If \(a\) is a positive real number, then the square roots of \(-a\) are \(\pm i \sqrt{a}\).

Worked exercise: Find the square roots of \(-4\).
Solution: The square roots of \(-4\) are the solutions to the equation $$\begin{align*}
z^2 &= -4\\
z^2 + 4 &= 0\\
z^2 -4i^2 &= 0 \\
(z-2i)(z+2i) &=0 \\
z &= 2i,-2i. \end{align*}$$ Alternatively, notice that \(-4 = 4i^2 = (2i)^2 = (-2i)^2\), which also shows us that the square roots are \(\pm 2i\).

We try to avoid writing \( \sqrt{-4} \) because it is ambiguous whether we are referring to \(2i\) or \(-2i\).

GL tip: We need to be careful when applying rules of real arithmetic to complex numbers. Some things that may be true for real numbers may not be true for complex numbers. For example, what went wrong here? $$1 = \sqrt{1} = \sqrt{(-1)\times(-1)} = \sqrt{-1} \times \sqrt{-1} = -1.$$

Square roots of a general complex number

The square roots of a complex number will also be complex numbers. The general method for finding the square root of a complex number \(w\) is:

  1. Let \(w = (x+yi)^2 = (x^2-y^2) + 2xyi\).
  2. Equate real and imaginary parts.
  3. Find \(x\) and \(y\) by solving simultaneously.

The general method of finding square roots is quite involved, so see if can find alternative methods before trying it!

Worked exercise: Find the square roots of \(3+4i\).
Solution: Let \(z = x+yi\) be the square root of \(3+4i\). Hence $$3+4i = (x+yi)^2 = (x^2-y^2)+2xyi.$$
Equating real and imaginary parts, we have:$$
\begin{align}
x^2-y^2 &=3 \label{1}\tag{1}\\
2xy &=4 \label{2}\tag{2}.
\end{align}$$
We need to solve these simultaneously. Using the normal method, you will end up with a quartic equation, so a more efficient way is to find \(x^2 + y^2\) using the following identity1: $$ (x^2 + y^2)^2 = (x^2-y^2)^2 + (2xy)^2.$$
Hence, we have \((x^2+y^2)^2 = 3^2+4^2=25\), and therefore $$x^2+y^2=5 \label{3} \tag{3}.$$ By adding \(\eqref{1}\) and \(\eqref{3}\), we have \(2x^2=8\), which solves to give us \(x=\pm2\).

Substituting this back to \(\eqref{2}\), gives us the solutions \((2,1)\) and \((-2,-1)\). (Note that substituting back to \(\eqref{1}\) is less useful because it will not give you which values of \(y\) corresponding to each value of \(x\).)

Hence, the square roots of \(3+4i\) are \(2+i\) and \(-2-i\).

GL tip: Faster advanced ways to find square roots
With some complex numbers, you can complete the square to find square roots in a couple of lines. For example, to find the square roots of \(3+4i\), we have $$3+4i = 4+4i + i^2 = (2+i)^2.$$ Hence, the square roots are \(\pm (2+i)\).

We can do this for many other complex numbers:
$$\begin{align*}
\pm 2i &= 1 \pm 2i + i^2 = (1\pm i)^2 \\
8 \pm 6i &= 9 \pm 6i + i^2 = (3 \pm i)^2 \\
15 \pm 8i &= 16 \pm 8i + i^2 = (4\pm i)^2.
\end{align*}$$
You can also use this trick with complex numbers derived from these. For example, to find the square roots of \(i\), we have
$$i = \frac{2i}{2} = \frac{1}{2} (1 + i)^2,$$ and hence the square roots are \( \pm \frac{1}{\sqrt{2}} (1+i)\).

To find the square roots of \(4+3i\), we have $$4+3i = i(3-4i)=i(4-4i+i^2)=i(2-i)^2.$$ Now we know the square roots of \(i\) are \( \pm \frac{1}{\sqrt{2}} (1+i)\), so the square roots of \(4+3i\) are \( \frac{1}{\sqrt{2}} (1+i)(2-i) = \frac{1}{\sqrt{2}} (3+i)\).

Problems

  1. Use the general method to find the square roots of:
    1. \(-9\)
    2. \(-10\)
    3. \(3-4i\)
    4. \(i\)
    5. \(8+6i\)
    6. \(8-6i\)
    7. \(5-12i\)
  2. Hard. Use a faster method to find the square roots of:
    1. \(-9\)
    2. \(-10\)
    3. \(3-4i\)
    4. \(i\)
    5. \(8+6i\)
    6. \(8-6i\)
  1. Alternatively, you can take the modulus of both sides of the equation \(3+4i = (x+yi)^2\).