# Question Video: Finding the Square Roots of Complex Numbers in Algebraic Form Mathematics

Without first converting π§ to trigonometric form, determine the square roots of π§, where π§ = (β8 + 2π)/(1 + 4π).

06:06

### Video Transcript

Without first converting π§ to trigonometric form, determine the square roots of π§, where π§ equals negative eight add two π over one add four π.

Before we begin trying to solve this problem, letβs write our complex number π§ in the form π plus ππ, where π is the real part and π is the imaginary part. To do this, weβll divide the complex number negative eight add two π by one add four π. We follow the usual procedure by using the conjugate of the denominator, that is, one minus four π. So we multiply π§ by one minus four π over one minus four π, which is essentially just multiplying by one. So weβre not changing π§ in any way.

Next, we multiply the numerators together. That gives us negative eight minus eight π squared plus two π plus 32π. And then, we multiply the denominators together. That gives us one minus 16π squared plus four π minus four π. But now remember that by definition, π squared is equal to negative one. So where we see π squared, we can replace this with negative one. And now, we can just simplify this a little bit. Starting with the numerator, eight multiplied by negative one is negative eight. But because weβve already got a minus at the front, this becomes a plus eight. And then, negative eight add eight is just going to give us zero.

We can also gather together like terms. Two π add 32π will give us 34π. Then for the denominator, 16 multiplied by negative one will give us negative 16. But because thereβs already a minus at the front, this is plus 16. And four π subtract four π will just give us zero. So we end up with 34π over 17. But this just gives us two π.

So now weβve rewritten π§ as two π, we can now proceed to find its square roots. Iβm just gonna clear some space so that we can do this. Letβs say the square root of two π equals the complex number π plus ππ. Weβre going to find the values for π and π. Starting off by squaring both sides gives us two π equals π squared add πππ add πππ add π squared π squared. We can then simplify this to π squared add two πππ add π squared π squared. But remember by definition, π squared is equal to negative one. So we can write this as minus π squared.

In order to solve this for π and π, we can group these terms together, like so. Then, we can compare the terms on the left of the expression with the terms on the right of the expression. On the left of the expression, we have no real terms. So we could say that the real part is zero. And on the right of the expression, the real part is π squared minus π squared. So this gives us that π squared minus π squared is equal to zero. And now comparing the imaginary part on the left with the imaginary part on the right, the imaginary part on the left is two. And the imaginary part on the right is two ππ. So that gives us that two ππ must be equal to two. But we can actually cancel this. Two ππ equals one.

So now, we have two expressions involving π and π. Weβre going to use these two expressions now to solve for π and π. From equation two, we have that π equals one over π. So letβs substitute this into equation one. This gives us π squared minus one over π squared equals zero. Letβs now evaluate one over π squared. This is going to give us π squared minus one over π squared equals zero, because we just square the numerator and we square the denominator.

To proceed from here, letβs multiply through by π squared. That gives us π to the fourth power minus one equals zero. And to solve for π, we can factor this expression as π squared add one multiplied by π squared minus one equals zero. This gives us the solutions of π squared equals negative one and π squared equals one. π squared equals one gives us that π is positive or negative one and π squared equals negative one gives us that π equals positive or negative π.

So now we found values for π, weβve got to find the corresponding values for π. Letβs take π to be positive or negative one. When π equals one, π equals one over one. This is just using the expression that we found earlier for π. And one over one is just one. When π equals negative one, π equals one over negative one, which is just negative one. So then, the complex number which we said is our solution to the square root of two π, π plus ππ. When π is one and π is one, this is one plus π. And when π is negative one and π is negative one, this is negative one minus π.

As weβre looking for a square root, weβre going to have two solutions. So these are our two solutions: one plus π and negative one minus π. But you might be wondering about the two other values we found for π, positive and negative π. Well actually, if we take π to be π, then π would be one over π. And if we take π to be negative π, then π would be one over negative π or just negative one over π.

When π equals π and π equals one over π, π plus ππ is π plus one over π multiplied by π. But thatβs just π plus one. And when π is negative π and π is negative one over π, π plus ππ is negative π plus negative one over π multiplied by π. But that just gives us negative π minus one. And actually, we see that these are exactly the same solutions that we just found. So the square roots of π§ are one plus π and negative one minus π.