Lesson Video: De Moivre’s Theorem | Nagwa Lesson Video: De Moivre’s Theorem | Nagwa

# Lesson Video: De Moivre’s Theorem Mathematics

In this video, we will learn how to find powers and roots of complex numbers and how to use de Moivre’s theorem to simplify calculations of powers and roots.

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### Video Transcript

In this video, we’ll learn how to use de Moivre’s theorem to simplify powers and roots of complex numbers. We’ll learn what the theorem states and look at where it comes from before using De Moivre’s theorem to solve problems involving powers and roots, including those involving other operations on complex numbers.

de Moivre’s theorem allows us to evaluate powers of a complex number written in polar form really quickly. We know that a complex number written in polar form can also be expressed in exponential form as 𝑟𝑒 to the 𝑖𝜃. We’re going to raise each side of this equation to the power of some integer value. Let’s call that 𝑛. So we can say that 𝑟𝑒 to the 𝑖𝜃 to the power of 𝑛 is equal to 𝑟 cos𝜃 plus 𝑖 sin𝜃 all to the power of 𝑛.

Since 𝑛 is an integer value, we can rewrite the left-hand side of this equation as 𝑟 to the power of 𝑛 times 𝑒 to the power of 𝑖𝑛𝜃. But this actually means that we can rewrite the right-hand side using Euler’s formula. And we get 𝑟 to the power of 𝑛 times cos 𝑛𝜃 plus 𝑖 sin 𝑛𝜃. And this is De Moivre’s theorem. For an integer value of 𝑛, 𝑟 cos𝜃 plus 𝑖 sin𝜃 to the power of 𝑛 is equal to 𝑟 to the power of 𝑛 times cos 𝑛𝜃 plus 𝑖 sin 𝑛𝜃.

It’s important to realize that this isn’t a strict proof of de Moivre’s theorem and that a formal proof comes from a more thorough understanding of the exponential form of a complex number. However, as previously mentioned, this theorem can allow us to easily evaluate powers of a complex number written in polar form. Let’s see what this might look like.

Simplify the square root of five times cos three 𝜋 by 14 plus 𝑖 sin of three 𝜋 by 14 to the power of seven times root three multiplied by cos of five 𝜋 by 22 plus 𝑖 sin of five 𝜋 by 22 to the power of 11.

In this question, we have the product of two complex numbers which are both written in polar form. To simplify these, we’ll need to use de Moivre’s theorem to help us evaluate the powers of each complex number before finding their product. Remember this theorem says that for integer values of 𝑛, a complex number written in polar form raised to the power of 𝑛 is equal to 𝑟 to the power of 𝑛 times cos 𝑛𝜃 plus 𝑖 sin 𝑛𝜃. Let’s use this to evaluate our first complex number to the power of seven.

For this number, 𝑟 is root five and 𝜃 is three 𝜋 by 14. We can rewrite this then as root five to the power of seven times cos of seven multiplied by three 𝜋 by 14 plus 𝑖 sin of seven multiplied by three 𝜋 by 14. We can simplify this and we see that the first complex number raised to the power of seven is 125 root five times cos of three 𝜋 by two plus 𝑖 sin of three 𝜋 by two. Similarly, for our second complex number, we raise the modulus which is root three to the power of 11. And we get 243 root three. And we multiply the argument — that’s five 𝜋 by 22 by 11 — and that gives us five 𝜋 by two.

Our final step is to find the product of these two complex numbers. To multiply complex numbers, we multiply their moduli and we add their arguments. 125 root five multiplied by 243 root three is 30375 root 15. And if we add their arguments, we get four 𝜋 and we can see that our complex number can be expressed as 30375 root 15 times cos four 𝜋 plus 𝑖 sin four 𝜋. Actually, we can simplify this a little bit further since cos of four 𝜋 is one and sin of four 𝜋 is zero. So our final answer is 30375 root 15.

In our next example, we’ll demonstrate how to use de Moivre’s theorem to simplify the quotient of powers of complex numbers.

Simplify 18 times negative 𝑖 plus one to the power of 39 divided by 𝑖 plus one to the power of 41.

Here we have the quotient of two complex numbers individually raised to the power of 39 and 41. We can use de Moivre’s theorem to evaluate these only when they’re in polar or exponential form. So let’s begin by writing negative 𝑖 plus one and 𝑖 plus one in polar form. To do this, we need to know the value of their moduli and arguments. The modulus is fairly straightforward. We calculate the square root of the sum of the squares of the real and imaginary parts of this number.

So the modulus of negative 𝑖 plus one is the square root of one squared plus negative one squared which is root two. Similarly, for 𝑖 plus one, that’s also root two. But what about their arguments? We’ll consider these individually. Negative 𝑖 plus one has a positive real part and a negative imaginary part. So it must lie in the fourth quadrant. We can therefore find its argument by using the formula the arctan of the imaginary divided by the real part. That’s the arctan of negative one divided by one which is negative 𝜋 by four.

𝑖 plus one lies in the first quadrant. So we can use the same formula. It’s the arctan of one divided by one which is 𝜋 by four. And we can see that we have our two complex numbers written in polar form. I’ve substituted them back into our fraction. What we’re going to need to do next is to evaluate the complex number on the top to the power of 39 and the one on the bottom to the power of 41.

Using de Moivre’s theorem, on the numerator, we have the square root of two to the power of 39 times cos 39 multiplied by negative 𝜋 by four plus 𝑖 sin of 39 multiplied by negative 𝜋 by four. And on the denominator, the modulus is root two to the power of 41 and the argument is 41 times 𝜋 by four. We don’t actually need to evaluate these yet. Instead, we recall the fact that to divide two complex numbers in polar form, we divide their moduli and subtract their arguments.

Dividing their moduli and we see that we’re left with 18 over root two squared, which is nine. Then, subtracting their arguments, we get an argument of negative 20𝜋. Now, cos of negative 20𝜋 is one and sin of negative 20𝜋 is zero. So we’re left with nine.

Now, you’ve probably spotted that we can generalize the properties of the modulus and the argument to integer powers of 𝑛. For a complex number 𝑧 and integer values of 𝑛, the modulus of 𝑧 to the power of 𝑛 is the same as the modulus of 𝑧 to the power of 𝑛. And the argument of 𝑧 to the power of 𝑛 is the same as 𝑛 times the argument of 𝑧.

And in fact, it’s also useful to know that we can generalize de Moivre’s theorem for a complex number and its conjugate. We don’t have time to demonstrate where this comes from in this video. But it’s useful to know that for the conjugate of 𝑧, which is denoted here by 𝑧 star, the conjugate of 𝑧 to the power of 𝑛 is equal to the conjugate of 𝑧 to the power of 𝑛. This formula can be really useful in helping us solve problems where we might not want to use De Moivre’s theorem fully. Let’s see what this might look like.

Given that 𝑧 is equal to root three minus 𝑖 to the power of 𝑛 and the modulus of 𝑧 is equal to 32, determine the principal argument of 𝑧.

The principal argument of 𝑧 is the value of 𝜃 such that 𝜃 is greater than negative 𝜋 and less than or equal to 𝜋. To answer this question, we’re going to recall the properties of the modulus. We know that the modulus of 𝑧 is equal to 32. So we can say that the modulus of root three minus 𝑖 to the power of 𝑛 is equal to 32. Using the properties of the modulus, we can rewrite this. And we can say that the modulus of root three minus 𝑖 all to the power of 𝑛 is equal to 32.

The modulus of root three minus 𝑖 is the square root of root three squared plus negative one squared. And that’s simply two. We can say then that two to the power of 𝑛 is equal to 32. And we know that two to the power of five is 32. So 𝑛 must be equal to five. We can now rewrite our complex number as root three minus 𝑖 all to the power of five. We then recall the rule that the argument of 𝑧 to the power of 𝑛 is equal to 𝑛 times the argument of 𝑧. This means that the argument of 𝑧 or the argument of root three minus 𝑖 to the power of five is five times the argument of root three minus 𝑖.

Now, root three minus 𝑖 lies in the fourth quadrant when plotted on the Argand diagram. And this is because its real part is positive and its imaginary part is negative. And we can find the argument of root three minus 𝑖 by using the formula the arctan of the imaginary part divided by the real part. That’s arctan of negative one over root three. That’s negative 𝜋 by six. So we can see that the argument of 𝑧 is five multiplied by negative 𝜋 by six which is negative five 𝜋 by six. This satisfies the criteria of being less than or equal to 𝜋 and greater than or equal to negative 𝜋. So we found the principal argument of 𝑧. It’s negative five 𝜋 by six.

For our final example, we’re going to look at how to use de Moivre’s theorem to find roots. De Moivre’s theorem for roots says that for a complex number 𝑟 cos 𝜃 plus 𝑖 sin 𝜃, its 𝑛 roots are 𝑟 to the power of one over 𝑛 times cos of 𝜃 plus two 𝜋 𝑘 over 𝑛 plus 𝑖 sin of 𝜃 plus two 𝑘 over 𝑛. And 𝑘 takes integer values from zero through to 𝑛 minus one.

Find the fourth roots of negative one, giving your answers in trigonometric form.

de Moivre’s theorem for roots says that the 𝑛th root of a complex number written in polar form is 𝑟 to the one over 𝑛 times cos of 𝜃 plus two 𝜋 𝑘 over 𝑛 plus 𝑖 sin of 𝜃 plus two 𝜋 𝑘 over 𝑛, where 𝑘 takes integer values of zero through to 𝑛 minus one. We therefore need to express negative one in polar form. The modulus of negative one is one and its argument is 𝜋. In polar form then, negative one is the same as cos 𝜋 plus 𝑖 sin 𝜋.

By applying de Moivre’s rule for roots, we can see that the fourth roots of negative one are cos of 𝜋 plus two 𝜋 𝑘 over four plus 𝑖 sin of 𝜋 plus two 𝜋 𝑘 over four, where 𝑘 takes values of zero through to three. We’ll begin by considering the root when 𝑘 is equal to zero. When 𝑘 is zero, our root is cos of 𝜋 by four plus 𝑖 sin of 𝜋 by four. When 𝑘 is one, our root is cos of 𝜋 plus two 𝜋 over four plus 𝑖 sin of 𝜋 plus two 𝜋 over four, which simplifies to cos of three 𝜋 by four plus 𝑖 sin of three 𝜋 by four.

When 𝑘 is two, we get cos of 𝜋 plus four 𝜋 over four plus 𝑖 sin of 𝜋 plus four 𝜋 over four. And that leaves us with an argument of five 𝜋 by four. But this argument is not in the range for the principal argument. Remember we can add or subtract multiples of two 𝜋 to achieve this. This time, we subtract two 𝜋 and we get an argument of negative three 𝜋 by four. So our third root is cos of negative three 𝜋 by four plus 𝑖 sin of negative three 𝜋 by four.

Then when 𝑘 is equal to three, the argument is 𝜋 plus six 𝜋 over four. This is seven 𝜋 over four, which once again is outside of the range for the principal argument. We subtract two 𝜋 and we get the principal argument for this root to be negative 𝜋 by four. So our fourth root is cos of negative 𝜋 by four plus 𝑖 sin of negative 𝜋 by four.

Now the reason we’ve stopped here is because de Moivre’s theorem for roots says that 𝑘 takes values of zero through to 𝑛 minus one. Aside from the fact that we know the fourth root of negative one will give us four roots, let’s have a look at a reason why we stop at minus one for 𝑘. Let’s look at what would have happened had we tried to evaluate the root when 𝑘 is equal to four.

We would have got cos of 𝜋 plus eight 𝜋 over four plus 𝑖 sin of 𝜋 plus eight 𝜋 over four. That would give us an argument of nine 𝜋 over four, which once again is outside the range for the principal argument. Subtracting two 𝜋 from nine 𝜋 over four and we get 𝜋 by four. Now, you should be able to see when 𝑘 is equal to four, we get the same result as if 𝑘 is equal to zero. So we just need the four values of 𝑘: zero, one, two, and three.

And we’ve used de Moivre’s theorem for roots to find the fourth roots of negative one. They are cos of 𝜋 by four plus 𝑖 sin 𝜋 by four, cos of three 𝜋 four plus 𝑖 sin of three 𝜋 by four, cos of negative three 𝜋 by four plus 𝑖 sin of negative three 𝜋 by four, and cos of negative 𝜋 by four plus 𝑖 sin of negative 𝜋 by four.

In this video, we’ve seen that de Moivre’s theorem says that we can raise a complex number written in polar form to an integer power of 𝑛 by using the formula 𝑟 to the power of 𝑛 times cos of 𝑛 𝜃 plus 𝑖 sin of 𝑛 𝜃. We’ve also seen that de Moivre’s theorem extends to finding 𝑛th roots of complex numbers. We evaluate 𝑟 to the power of one over 𝑛 times cos of 𝜃 plus two 𝜋 𝑘 over 𝑛 plus 𝑖 sin of 𝜃 plus two 𝜋 𝑘 over 𝑛, where 𝑘 takes integer values from zero through to 𝑛 minus one.

And of course, we’ve explicitly stated throughout this video the de Moivre’s theorem is used to calculate 𝑛th roots and integer powers. But we can’t actually assume that it applies to a real or complex exponents.