Video: Finding Powers of Complex Numbers in Polar Form

Given that 𝑧 = 2√3(cos 240Β° + 𝑖 sin 240Β°), find 𝑧² in exponential form.

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Video Transcript

Given 𝑧 is equal to two root three multiplied by cos of 240 degrees plus 𝑖 sin of 240 degrees, find 𝑧 squared in exponential form.

We’re currently given a complex number written in trigonometric form. And we’re looking to find 𝑧 squared in exponential form. There are two ways we can go about this. We can evaluate 𝑧 squared in trigonometric form, and then convert it to exponential form. Or we can convert it to exponential form first, and then work out the value of 𝑧 squared. Let’s consider both of these methods.

And to square this complex number, we recall De Moivre’s theorem. And this said, for a complex number in trigonometric form π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ, this complex number to the power of 𝑛 is given by π‘Ÿ to the power of 𝑛 multiplied by cos π‘›πœƒ plus 𝑖 sin π‘›πœƒ. And in this example, 𝑛 is a natural number.

We can see that the modulus of our complex number 𝑧 is two root three. And πœƒ, its argument is 240 degrees. In fact, at some point, we’re going to have to convert this to radians. So, we might as well do that now and get it out of the way. To do this, we recall the fact that two πœ‹ radians is equal to 360 degrees. And we can find the value of one degree by dividing through by 360. One degree is equal to two πœ‹ over 360 radians. And two πœ‹ by 360 simplifies to πœ‹ by 180. So, one degree is equal to πœ‹ over 180 radians. So, we can change 240 degrees into radians by multiplying it by πœ‹ over 180. That gives us four πœ‹ by three.

And so, we can work out the modulus of 𝑧 squared by squaring the modulus of 𝑧. That’s two root three squared. Root three squared is three. So, two root three squared is two squared multiplied by three, which is 12. And then, to work out the argument of 𝑧 squared, we multiply the argument of 𝑧 by the power that’s two. Four πœ‹ by three multiplied by two is eight πœ‹ by three. So, we can see that in trigonometric form, 𝑧 squared is 12 multiplied by cos of eight πœ‹ over three plus 𝑖 sin of eight πœ‹ over three.

And remember, to change a complex number in trigonometric form into exponential form, it’s π‘Ÿπ‘’ to the π‘–πœƒ. And since π‘Ÿ for 𝑧 squared the modulus is 12 and the argument πœƒ is eight πœƒ by three, we can say that 𝑧 squared is equal to 12𝑒 to the eight πœ‹ over three 𝑖. Remember though, we usually want to represent this using the principal argument. That’s greater than negative πœ‹ and less than or equal to πœ‹. In fact, eight πœ‹ by three is greater than πœ‹. So, to find the principal argument, we add or subtract multiples of two πœ‹.

Here, let’s subtract two πœ‹ from eight πœ‹ by three. Two πœ‹ is equal to six πœ‹ over three. And when we subtract six πœ‹ over three from eight πœ‹ over three, we’re left with two πœ‹ over three. So, in exponential form, 𝑧 squared is 12𝑒 to the two πœ‹ by three 𝑖.

Now, let’s consider the alternative method. And that was to convert this complex number into exponential form first and then square it. Once again, we’ll use this rule. A complex number with a modulus of π‘Ÿ and an argument πœƒ can be represented in exponential form as π‘Ÿπ‘’ to the π‘–πœƒ. We already solved 240 degrees is equal to four πœ‹ by three radians. So, we can say that 𝑧 in exponential form is two root three multiplied by 𝑒 to the power of four πœ‹ by three 𝑖.

And this time, to find 𝑧 squared, we consider the alternative form of De Moivre’s theorem. And that says that if 𝑧 is equal to π‘Ÿπ‘’ to the π‘–πœƒ, then 𝑧 to the power of 𝑛 is equal to π‘Ÿ to the power of 𝑛 multiplied by 𝑒 to the π‘–π‘›πœƒ. And now you should be able to see the relationship between the two forms and the methods that we’re using.

This time, 𝑧 squared is equal to two root three squared multiplied by 𝑒 to the power of two multiplied by four πœ‹ by three 𝑖. And, once again, we know that two root three squared is 12. And two multiplied by four πœ‹ by three is eight πœ‹ by three. And, once again, changing our argument into the principal argument by subtracting two πœ‹, we can see that 𝑧 squared is equal to 12𝑒 to the two πœ‹ by three 𝑖.

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