# Question Video: Finding the Square Roots of Complex Numbers Using De Moivre’s Theorem

Use De Moivre’s theorem to find the two square roots of 9(cos (2𝜋/3) + 𝑖 sin (2𝜋/3)).

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### Video Transcript

Use De Moivre’s theorem to find the two square roots of nine times the cos of two 𝜋 by three plus 𝑖 sin of two 𝜋 by three.

In this question, we’re given a complex number in trigonometric form, and we need to use this to determine both of the values of the two square roots of this number. To do this, we need to use De Moivre’s theorem.

To do this, let’s start by recalling De Moivre’s theorem for roots of complex numbers. This tells us if 𝑍 is a complex number in trigonometric form, that’s 𝑟 times the cos of 𝜃 plus 𝑖 sin of 𝜃, then we can find all of the 𝑛th roots of 𝑍 by using the following formula. They’re given by 𝑟 to the power of one over 𝑛 times the cos of 𝜃 plus two 𝜋𝑘 over 𝑛 plus 𝑖 sin of 𝜃 plus two 𝜋𝑘 over 𝑛, where 𝑘 is any integer between zero and 𝑛 minus one inclusive.

We want to find the square roots of our number. So our value of 𝑛 is going to be two, since the square roots are the second roots. And if we call the complex number given to us in the question 𝑍, we can see that 𝑍 is already given in trigonometric form. We can see that our value of 𝑟, the modulus of 𝑍, is nine and the value of 𝜃, the argument of 𝑍, is equal to two 𝜋 by three. We can substitute these values into our formula to find the two square roots of this number.

Substituting these values in, we get nine to the power of one-half multiplied by the cos of two 𝜋 by three plus two 𝜋𝑘 all over two plus 𝑖 sin of two 𝜋 by three plus two 𝜋𝑘 all over two. And this gives us a root for each possible value of 𝑘. Remember, 𝑘 is an integer which varies from zero to 𝑛 minus one. In this case, since 𝑛 is two, 𝑘 can be either zero or one.

So let’s substitute both of these values into this expression. Let’s start with 𝑘 is equal to zero. First, we have nine to the power of one-half, which we know is equal to three. Next, when 𝑘 is equal to zero, two 𝜋 multiplied by 𝑘 is equal to zero. So we have the cos of two 𝜋 by three all over two, which is the cos of 𝜋 by three. And the exact same will be true in our second sum. We have 𝑖 sin of 𝜋 by three. This then gives us an expression for the first square root of our complex number 𝑍: three multiplied by the cos of 𝜋 by three plus 𝑖 times the sin of 𝜋 by three.

We can then evaluate the cos of 𝜋 by three, which is one-half, and the sin of 𝜋 by three, which is root three over two, and then distribute three over our parentheses to find the first root of our complex number. It’s equal to three over two plus three root three over two 𝑖. We can then find the other square root of our complex number 𝑍 by substituting 𝑘 is equal to one into this expression. This time, when we substitute 𝑘 is equal to one and we evaluate our argument, we get two 𝜋 by three plus two 𝜋 all divided by two, which we evaluate is equal to four 𝜋 by three. Therefore, the other square root of this number is given by three multiplied by the cos of four 𝜋 by three plus 𝑖 sin of four 𝜋 by three.

And if we evaluate the cosine and sine in this expression and distribute three over the parentheses, we get negative three over two minus three root three over two 𝑖. We’ll write these two roots in a set to get our final answer. Therefore, by using De Moivre’s theorem, we were able to find the two square roots of nine times the cos of two 𝜋 by three plus 𝑖 sin of two 𝜋 by three. They are the two elements of the set containing three over two plus three root three over two 𝑖 and negative three over two minus three root three over two 𝑖.