Question

find the square root of the following complex number21-20i​

Answer 1

Answer:

Refer to the attachment.....

Hope it helps you.....

Answer 2

Answer:

We know that all square roots of this number will satisfy the equation 21-20i=x² 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=a²+(2ab)i+(b²)i²

As i²=-1 by definition of i, this equation can be rearranged to give 21-20i=(a²-b²)+(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 a²-b²=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 a²-(-10/a)²=21.

Some simplification and factorisation of this equation gives us (a²+4)(a²-25)=0, a quadratic in disguise.

So either a²=-4 or a²=25.

We have assumed a to be real so a²=-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.