# How do you find the square roots of a complex number?

*Every complex number has complex square roots. However since we don't know how to deal with expressions such as √i we need to follow a specific method to find the square roots of a complex number.*

**Let's consider the complex number 21-20i.**

We know that all square roots of this number will satisfy the equation 21-20i=x^{2} by definition of a square root.

We also know that x can be expressed as a+bi (where a and b are real) since the square roots of a complex number are always complex.

**So 21-20i=(a+bi) ^{2}.**

The natural step to take here is the mulitply out the term on the right-hand side.

This gives 21-20i=a^{2}+(2ab)i+(b^{2})i^{2}.

As i^{2}=-1 by definition of i, this equation can be rearranged to give 21-20i=(a^{2}-b^{2})+(2ab)i.

Now both sides of the equation are in the same form.

**Let's compare coeffiecients to obtain two equations in a and b.**

First, let's compare the real parts of the equation.

We have a^{2}-b^{2}=21 (call this equation 1).

Next, let's compare the imaginary parts of the equation (the coefficients of i).

We have 2ab=-20 (call this equation 2).

**We now have two equations in two unknowns. We can solve these simultaneous equations for a and b.**

Firstly, we can make b the subject of equation 2 by dividing both sides by 2a.

We have b=-10/a.

Now substitute this expression for b into equation 1.

We have a^{2}-(-10/a)^{2}=21.

Some simplification and factorisation of this equation gives us (a^{2}+4)(a^{2}-25)=0, a quadratic in disguise.

So either a^{2}=-4 or a^{2}=25.

We have assumed a to be real so a^{2}=-4 has no solutions of interest to us.

This means our solutions are a=5 and a=-5.

Substitute each a value into our earlier expression for b.

**This means that when a=5, b=-2 and when a=-5, b=2.**

**So, putting a and b back into the context of the question, we have two solutions: 5-2i and -5+2i.**