Square Roots of a Complex Number

The square roots of a complex number will themselves be complex:
- i.e. if $z^{2} = a + b i$ then $z = c + d i$
We can then square ( $c + d i$ ) and equate it to the original complex number ( $a + b i$ ), as they both describe $z^{2}$ :
- $a + b i = {(c + d i)}^{2}$
Then expand and simplify:
- $a + b i = c^{2} + 2 c d i + d^{2} i^{2}$
- $a + b i = c^{2} + 2 c d i - d^{2}$
As both sides are equal we are able to equate real and imaginary parts:
- Equating the real components: $a = c^{2} - d^{2}$ (1)
- Equating the imaginary components: $b = 2 c d$ (2)
These equations can then be solved simultaneously to find the real and imaginary components of the square root
- In general, we can rearrange (2) to make $\frac{b}{2 d} = c$ and then substitute into (1)
- This will lead to a quartic equation in terms of d; which can be solved by making a substitution to turn it into a quadratic (see 1.1.5 Further Solving Quadratic Equations (Hidden Quadratics))
The values of $d$ can then be used to find the corresponding values of $c$ , so we now have both components of both square roots ( $c + d i$ )
Note that one root will be the negative of the other root
- i.e. $c + d i$ and $- c - d i$

8-1-3-square-root-of-complex-number-part-1

8-1-3-square-root-of-complex-number-part-2

Most calculators used at A-Level can handle complex numbers.
Once you have found the square roots algebraically; use your calculator to square them and make sure you get the number you were originally trying to square-root!

Square Roots of a Complex Number (CIE A Level Maths: Pure 3)