Question Video: Finding the Power of the Product of Complex Numbers in Exponential Form | Nagwa Question Video: Finding the Power of the Product of Complex Numbers in Exponential Form | Nagwa

Question Video: Finding the Power of the Product of Complex Numbers in Exponential Form Mathematics • Third Year of Secondary School

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If 𝑍₁ = 6(cos 225Β° + 𝑖 sin 225Β°), 𝑍₂ = cos 90Β° + 𝑖 sin 90Β°, and 𝑍₃ = cos 270Β° + 𝑖 sin 270Β°, what is the exponential form of (𝑍₁𝑍₂ 𝑍₃)Β²?

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Video Transcript

If 𝑍 sub one is equal to six times cos of 225 degrees plus 𝑖 sin of 225 degrees, 𝑍 sub two is equal to cos of 90 degrees plus 𝑖 sin of 90 degrees, and 𝑍 sub three is equal to cos of 270 degrees plus 𝑖 sin of 270 degrees, what is the exponential form of 𝑍 sub one times 𝑍 sub two times 𝑍 sub three squared?

In this question, we’ve been given three complex numbers represented in polar or trigonometric form. We’re being asked to find the product of these numbers, that’s 𝑍 sub one times 𝑍 sub two times 𝑍 sub three. And then we’re going to raise that to a power of two. And in fact, we’re going to give our answer in exponential form.

In fact, converting between polar and exponential form is fairly straightforward. For a complex number of the form 𝑍 equals π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ, its equivalent exponential form is π‘Ÿπ‘’ to the π‘–πœƒ. Here, π‘Ÿ is known as the modulus, whilst πœƒ is the argument. And when we work in exponential form, we ensure that our argument is actually in terms of radians.

There are two ways that we can find the products of these complex numbers and then square that value. We could first multiply each of the numbers together in polar form, find their square, and then convert it. Or we could begin by converting it into exponential form and then finding the product and squaring it. We’re going to use the first method; we’re going to find the product of our numbers in polar form.

We recall that multiplying complex numbers in polar form is really straightforward. We simply multiply their moduli and add their arguments. So if π‘Ÿ sub one, π‘Ÿ sub two, and π‘Ÿ sub three are the moduli of our complex numbers, then we need to find the product of these. We see that the modulus of 𝑍 sub one is six. The modulus of 𝑍 sub two is in fact one. And similarly, the modulus of 𝑍 sub three is also one. This means that our new modulus of 𝑍 sub one times 𝑍 sub two times 𝑍 sub three is just going to be equal to six.

Then the argument of the product of our complex numbers will be the sum of their respective arguments. Their arguments are 225 degrees, 90 degrees, and 270 degrees. And so we find the sum of these, giving us a value of 585 degrees.

Now, in fact, we are going to eventually convert this into exponential form. So at this stage, it makes sense to convert this into radian measure. We might recall that πœ‹ radians is equal to 180 degrees. Or equivalently, by dividing both sides of this equation by 180, we find that one degree is equal to πœ‹ over 180 radians. And so our argument is the product of 585 and πœ‹ over 180 radians. That’s 13πœ‹ by four radians.

And so we’re now able to write the complex number 𝑍 sub one 𝑍 sub two 𝑍 sub three as six times cos of 13πœ‹ by four plus 𝑖 sin of 13πœ‹ by four.

Now remember we’re not quite finished. We’re looking to square this value. To do this, we’re going to use de Moivre’s theorem. And this says that if 𝑍 is equal to π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ, then 𝑍 raised to the 𝑛th power is π‘Ÿ to the 𝑛th power times cos of π‘›πœƒ plus 𝑖 sin of π‘›πœƒ. And that’s for integer values of 𝑛. We see then that we need to square the modulus of our complex number and double the argument as shown. This gives us a modulus of six squared, which is 36, and an argument of two times 13πœ‹ over four, which is 13πœ‹ by two.

Now at this stage, we’re almost ready to convert this into exponential form. However, before we do, we’re going to choose to write this in terms of the principal argument. In other words, we’re going to add or subtract multiples of two πœ‹ from our argument so that it’s in the range πœƒ is greater than negative πœ‹ and less than or equal to πœ‹. In fact, if we subtract six lots of two πœ‹, we end up with an argument of πœ‹ by two, which is in the range of the principal argument.

And so our new complex number is 36 times cos πœ‹ by two plus 𝑖 sin πœ‹ by two. And all that’s left is to convert this now into exponential form. To do so, we go back to the very first definitions we provided. We see that π‘Ÿ, the modulus of our new complex number, is 36. The argument is πœ‹ by two. And so we can put that back into the exponential form of a complex number.

Given the information about 𝑍 sub one, 𝑍 sub two, and 𝑍 sub three then, the exponential form of the product of these complex numbers all squared is 36 times 𝑒 to the πœ‹ by two 𝑖.

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