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 16(cos (5πœ‹/3) βˆ’ 𝑖 sin (5πœ‹/3)).

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Video Transcript

Use De Moivre’s theorem to find the two square roots of 16 times the cos of five πœ‹ by three minus 𝑖 times the sin of five πœ‹ by three.

We’re given a complex number in trigonometric form, which we’ll call 𝑧, and that’s equal to 16 times the cos of five πœ‹ by three minus 𝑖 sin five πœ‹ by three. And we’re asked to use De Moivre’s theorem to find the square root of 𝑧. Let’s just remind ourselves of what this says. For integer values of 𝑛, if 𝑧 is π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ, then the 𝑛th root of 𝑧 β€” that’s 𝑧 to the power one over 𝑛 β€” is π‘Ÿ to the power one over 𝑛 times the cos of πœƒ plus two π‘˜πœ‹ over 𝑛 plus 𝑖 sin πœƒ plus two π‘˜πœ‹ over 𝑛. And that’s for values of π‘˜ from zero to 𝑛 minus one.

Now comparing our 𝑧 to the one shown in De Moivre’s theorem, the first thing to notice is that we have a negative sine in between the cosine and the sine, whereas in De Moivre’s theorem this is positive. And to use De Moivre’s theorem, we’re going to use the fact that cos πœƒ is cos of negative πœƒ and negative sin πœƒ is the sin of negative πœƒ. So that our 𝑧 is actually equal to 16 times the cos of negative five πœ‹ by three plus 𝑖 times the sin of negative five πœ‹ by three. And our trigonometric expression is in the correct form to use De Moivre’s theorem. Since we’re looking for the square roots of our 𝑧, our 𝑛 is equal to two. And since π‘˜ takes values from zero to 𝑛 minus one, this then means that we have two values for π‘˜; that’s zero and one.

And now comparing our 𝑧 to the 𝑧 in De Moivre’s theorem, we have π‘Ÿ is equal to 16 and πœƒ is equal to negative five πœ‹ by three. So now, using De Moivre’s theorem, we have 𝑧 to the power of a half is equal to 16 to the power of a half times the cos of negative five πœ‹ by three plus two π‘˜πœ‹ all over two plus 𝑖 times the sin of negative five πœ‹ by three plus two π‘˜πœ‹ over two.

And all we need to do now is to evaluate this for our two values of π‘˜; that’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œ‹ by six plus 𝑖 times the sin of negative five πœ‹ by six. That’s because when π‘˜ is equal to zero, two π‘˜πœ‹ is also zero. And negative five πœ‹ divided by three all divided by two is equal to negative five πœ‹ by six. Now, the cos of negative five πœ‹ by six is negative root three over two, where the sin of negative five πœ‹ by six is negative a half. So that our first root is four times negative root three over two minus 𝑖 over root two. And evaluating this means our first root of 𝑧 where π‘˜ is equal to zero is negative two times the square root of three plus two 𝑖.

And now let’s make some room. And our second root where π‘˜ is equal to one is four times the cos of negative five πœ‹ by three plus two πœ‹ all over two plus 𝑖 times the sin of negative five πœ‹ by three plus two πœ‹ over two. That is four times the cos of πœ‹ by six plus 𝑖 times the sin of πœ‹ by six. The cos of πœ‹ by six is root three over two, and the sin of πœ‹ by six is one-half. So that our second root is four times the square root of three over two plus 𝑖 over two. That is two times the square root of three plus two 𝑖.

The two square roots of 16 times the cos of five πœ‹ by three minus 𝑖 times sin five πœ‹ by three are therefore plus or minus two times the square root of three plus two 𝑖.

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