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Question Video: Finding the Power of Complex Numbers in Exponential Form Mathematics

Given that 𝑧 = (√(3)/2) βˆ’ (3/2)𝑖, find 𝑧⁡, giving your answer in exponential form.

03:57

Video Transcript

Given that 𝑧 is equal to the square root of three over two minus three over two 𝑖, find 𝑧 to the fifth power, giving your answer in exponential form.

In this question, we’re given a complex number 𝑧 in algebraic form. We need to find 𝑧 to the fifth power, where we give our answer in exponential form. To do this, let’s start by recalling what we mean by the exponential form of a complex number. It’s the form π‘Ÿ times 𝑒 to the π‘–πœƒ, where π‘Ÿ is the modulus of the complex number and πœƒ is its argument. We need to write 𝑧 to the fifth power in this form. And to do this, we can recall a result involving integer exponent powers of complex numbers in exponential form. We know for any integer value of 𝑛, π‘Ÿ times 𝑒 to the power of π‘–πœƒ all raised to the 𝑛th power is equal to π‘Ÿ to the 𝑛th power times 𝑒 to the power of π‘–π‘›πœƒ. This is an application of de Moivre’s theorem, where π‘Ÿπ‘’ to the π‘–πœƒ is a complex number given in exponential form.

Since the exponent of five is an integer, we can use this to answer our question. We just need to write 𝑧 in exponential form. To write a complex number in exponential form, we need to find its modulus and its argument. Let’s start by finding the modulus of 𝑧. And we recall the modulus of a complex number is the square root of the sum of the squares of its real and imaginary parts. And we can find these values from the question. The real part of 𝑧 is root three over two, and the imaginary part of 𝑧 is negative three over two. So the modulus of 𝑧 is the square root of root three over two all squared plus negative three over two all squared. We can then simplify this. Root three over two all squared is three-quarters, and negative three over two all squared is nine over four. Then, we can just evaluate this expression. Three-quarters plus nine-quarters is twelve-quarters, which simplifies to give us three. So the modulus of 𝑧 is the square root of three.

So the square root of three is the value of π‘Ÿ in the exponential form of 𝑧. However, we still need to find the argument of 𝑧. And to find the argument of 𝑧, we first need to notice the real part of 𝑧 is positive but the imaginary part of 𝑧 is negative, which tells us in an Argand diagram 𝑧 lies in the fourth quadrant. This then allows us to find the value of πœƒ, the argument of 𝑧. We recall for any complex number in the first or fourth quadrant, the argument value πœƒ is the inverse tangent of its imaginary part divided by its real part. So, in our case, πœƒ is the inverse tan of negative three over two divided by root three over two.

We want to evaluate this expression. First, we can cancel the shared factor of one-half in the numerator and denominator. This gives us the inverse tan of negative three over root three. We can then simplify this by rationalizing the denominator. We get the inverse tan of negative three root three over three. Then, we cancel the shared factor of three in the numerator and denominator. We get the inverse tan of negative root three, which we can calculate is equal to negative πœ‹ by three. Therefore, we found the modulus of 𝑧 and the argument of 𝑧. So we can use this to write 𝑧 in exponential form. 𝑧 is root three times 𝑒 to the power of negative 𝑖 πœ‹ by three.

Now, we’re ready to use our result to find 𝑧 to the fifth power. First, we raise both sides of this expression to the fifth power. Next, since the exponent value of five is an integral, we can use our result to distribute the exponent over our parentheses. We get root three all raised to the fifth power multiplied by 𝑒 to the power of negative 𝑖 times five πœ‹ by three. We can then evaluate this expression. First, root three all raised to the fifth power is root three all squared multiplied by root three, which we can calculate is equal to nine root three. And this is enough to answer our question. However, we can simplify this slightly. If we call the argument of this number πœƒ one, we see that it’s the coefficient of 𝑖 in the exponent. That’s negative five πœ‹ by three. But remember, we can add and subtract integer multiples of two πœ‹ to the argument. And we’ll get an equivalent argument for our number. This then gives us πœ‹ by three.

So we found the exponential form of 𝑧 to the fifth power. It’s nine root three multiplied by 𝑒 to the power of πœ‹ by three times 𝑖.

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