# Video: Finding the Square Roots of Complex Numbers Using De Moivreβs Theorem

Use De Moivreβs theorem to find the two square roots of 16(cos (5π/3) β π sin (5π/3)).

03:22

### Video Transcript

Use De Moivreβs theorem to find the two square roots of 16 times the cos of five π by three minus π times the sin of five π by three.

Weβre given a complex number in trigonometric form, which weβll call π§, and thatβs equal to 16 times the cos of five π by three minus π sin five π by three. And weβre asked to use De Moivreβs theorem to find the square root of π§. Letβs just remind ourselves of what this says. For integer values of π, if π§ is π cos π plus π sin π, then the πth root of π§ β thatβs π§ to the power one over π β is π to the power one over π times the cos of π plus two ππ over π plus π sin π plus two ππ over π. And thatβs for values of π from zero to π minus one.

Now comparing our π§ to the one shown in De Moivreβs theorem, the first thing to notice is that we have a negative sine in between the cosine and the sine, whereas in De Moivreβs theorem this is positive. And to use De Moivreβs theorem, weβre going to use the fact that cos π is cos of negative π and negative sin π is the sin of negative π. So that our π§ is actually equal to 16 times the cos of negative five π by three plus π times the sin of negative five π by three. And our trigonometric expression is in the correct form to use De Moivreβs theorem. Since weβre looking for the square roots of our π§, our π is equal to two. And since π takes values from zero to π minus one, this then means that we have two values for π; thatβs zero and one.

And now comparing our π§ to the π§ in De Moivreβs theorem, we have π is equal to 16 and π is equal to negative five π by three. So now, using De Moivreβs theorem, we have π§ to the power of a half is equal to 16 to the power of a half times the cos of negative five π by three plus two ππ all over two plus π times the sin of negative five π by three plus two ππ over two.

And all we need to do now is to evaluate this for our two values of π; thatβs π is equal to zero and π is equal to one. And since the square root of 16 is four, when π is equal to zero, we have four times the cos of negative five π by six plus π times the sin of negative five π by six. Thatβs because when π is equal to zero, two ππ is also zero. And negative five π divided by three all divided by two is equal to negative five π by six. Now, the cos of negative five π by six is negative root three over two, where the sin of negative five π by six is negative a half. So that our first root is four times negative root three over two minus π over root two. And evaluating this means our first root of π§ where π is equal to zero is negative two times the square root of three plus two π.

And now letβs make some room. And our second root where π is equal to one is four times the cos of negative five π by three plus two π all over two plus π times the sin of negative five π by three plus two π over two. That is four times the cos of π by six plus π times the sin of π by six. The cos of π by six is root three over two, and the sin of π by six is one-half. So that our second root is four times the square root of three over two plus π over two. That is two times the square root of three plus two π.

The two square roots of 16 times the cos of five π by three minus π times sin five π by three are therefore plus or minus two times the square root of three plus two π.