Question Video: Complex Cubic Roots of Unity | Nagwa Question Video: Complex Cubic Roots of Unity | Nagwa

Question Video: Complex Cubic Roots of Unity Mathematics

Let 𝑧_(1) = 𝑒^(𝑖 2πœ‹/3) and 𝑧_2 = 𝑒^(βˆ’π‘– 2πœ‹/3) be the complex cubic roots of unity. 1. Evaluate 𝑧_(1)Β². How does this compare with 𝑧_2? 2. Evaluate 𝑧_(2)Β². How does this compare with 𝑧_1?

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Video Transcript

Let 𝑧 one equal 𝑒 to the 𝑖 two πœ‹ by three and 𝑧 two equal 𝑒 to the negative 𝑖 two πœ‹ by three be the complex cubic roots of unity. 1) Evaluate 𝑧 one squared. How does this compare with 𝑧 two? 2) Evaluate 𝑧 two squared. How does this compare with 𝑧 one?

Notice how 𝑧 one and 𝑧 two are the complex solutions to 𝑧 cubed equals one, written in exponential form. This means we can use De Moivre’s theorem to evaluate 𝑧 one squared. This says that, for a complex number of the form 𝑧 is equal to π‘Ÿπ‘’ to the π‘–πœƒ, 𝑧 to the power of 𝑛 is equal to π‘Ÿ to the power of 𝑛 times 𝑒 to the π‘–π‘›πœƒ. Remember, π‘Ÿ is the modulus and πœƒ is the argument. We can see that the modulus of 𝑧 one is simply one. And the argument of 𝑧 one is two πœ‹ by three. So 𝑧 one squared is one squared times 𝑒 to the 𝑖 two πœ‹ by three times two.

One squared is one. And two πœ‹ by three multiplied by two is four πœ‹ by three. So we can see that 𝑧 one squared is equal to 𝑒 to the 𝑖 four πœ‹ by three. The argument of 𝑧 one squared is outside of the range for the principal argument. So we’ll subtract two πœ‹. And we find that the principal argument of 𝑧 one squared is negative two πœ‹ by three. And we can now see that 𝑧 one squared is equal to 𝑧 two.

Let’s repeat this process for question two. Let’s begin by making a prediction. We saw that 𝑧 one squared is equal to 𝑧 two. So it might seem to follow that 𝑧 two squared will be equal to 𝑧 one. But let’s check. Once again, the modulus of 𝑧 two is one. But this time, its argument is negative two πœ‹ by three. Negative two πœ‹ by three multiplied by two is negative four πœ‹ by three. Once again, the argument of 𝑧 two squared is outside of the range for the principal argument.

This time, we’ll add two πœ‹. Remember, we’re allowed to add or subtract any multiple of two πœ‹ to achieve an argument that’s within the range for the principal argument. This time, the argument of 𝑧 two squared is two πœ‹ by three. And we now see that 𝑧 two squared is equal to 𝑧 one as we predicted. So 𝑧 one squared is equal to 𝑧 two. And 𝑧 two squared is equal to 𝑧 one, where 𝑧 one and 𝑧 two are the complex cubic roots of unity.

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