Video: Multiplication and Division of Complex Numbers Involving Polar and Exponential Forms

Given that 𝑧₁ = 8(cos 240Β°+ 𝑖 sin 240Β°), 𝑧₂ = 4(cos (5πœ‹/4) + 𝑖 sin (5πœ‹/4)), and 𝑧₃ = 8(cos 45Β°+ 𝑖 sin 45Β°), find (𝑧₁𝑧_(2) ^(6))/(𝑧_(3) ^(4)), giving your answer in exponential form.

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Video Transcript

Given that 𝑧 one is equal to eight times cos of 240 degrees plus 𝑖 sin of 240 degrees. 𝑧 two is equal to four times cos of five πœ‹ over four plus 𝑖 sin of five πœ‹ over four. And 𝑧 three is equal to eight times cos of 45 degrees plus 𝑖 sin of 45 degrees. Find 𝑧 one multiplied by 𝑧 two to the power of six divided by 𝑧 three to the power of four, giving your answer in exponential form.

Now for this question, we’ve been given three complex numbers. And we have to perform various operations, such as multiplication, division, and raising to some powers. We could approach this using De Moivre’s theorem. However, you may notice that the question asked for an answer in exponential form. Because of this, the method that we’re going to explore will first be to convert our complex numbers to exponential form and only then to perform the operations.

Let us first recall the trigonometric and exponential form for a complex number. You’ll notice that, in both of these forms, the complex number 𝑧 is expressed in terms of two main parameters: the magnitude π‘Ÿ and the angle πœƒ. We’re able to move between trigonometric and exponential form by taking the values of π‘Ÿ and πœƒ from one expression and inputting them into the other expression.

Before we do this, let us first align the units of πœƒ. Looking at our complex numbers, we can see that πœƒ is expressed in terms of degrees for 𝑧 one and 𝑧 three and in radians for 𝑧 two. We can convert all of our degree values to radians by remembering the following relationship. 360 degrees is two πœ‹ radians. This allows us to calculate that 240 degrees is four πœ‹ over three radians. And 90 degrees is πœ‹ over four radians.

Now that we found these values, let us rewrite our complex numbers with πœƒ expressed in terms of radians. 𝑧 one is equal to eight times cos of four πœ‹ over three plus 𝑖 sin of four πœ‹ over three. 𝑧 two remains unchanged, since we already had πœƒ in terms of radians. And 𝑧 three is equal to eight times cos of πœ‹ over four plus 𝑖 sin of πœ‹ over four.

Now that we’ve done this, let’s work on converting our three complex numbers from trigonometric form to exponential form. Remember, this can be achieved by taking the parameters of π‘Ÿ and πœƒ and inputting them into this expression. Starting with 𝑧 one, we can see that our value for π‘Ÿ is eight, and our value for πœƒ is four πœ‹ over three. 𝑧 one is therefore equal to eight times 𝑒 to the power of four πœ‹ over three times 𝑖.

Going through the same process for 𝑧 two, we find that this is equal to four times 𝑒 to the power of five πœ‹ over four times 𝑖. And finally, we find that 𝑧 three is equal to eight times 𝑒 to the power of πœ‹ over four times 𝑖.

Now that we have our three complex numbers in exponential form, let’s look at the operations that we’re going to be performing. Here we see that we have 𝑧 one multiplied by 𝑧 two to the power of six divided by 𝑧 three to the power of four. We can make our future calculations slightly easier by remembering that one over some number π‘₯ to the power of π‘Ž is equal to π‘₯ to the power of negative π‘Ž. This allows us to rewrite our question as 𝑧 one multiplied by 𝑧 two to the power of six multiplied by 𝑧 three to the power of negative four.

First, let’s tidy things up by getting rid of the trigonometric form of our complex numbers. We’ll leave some gaps to make room for our future calculations. Now let’s work on applying the powers to our complex numbers.

Our first complex number, 𝑧 one, has no power to apply. However, we’re gonna choose to express the multiple of eight in terms of powers of two. The reasons for this will become apparent later. Eight is equal to two to the power of three. We can also do the same for 𝑧 two and 𝑧 three, finding that four is equal to two squared and eight once again is equal to two to the power of three.

Let’s sub in these values. We now need to raise 𝑧 two to the power of six. This can be done by raising two squared to the power of six, which is multiplied by 𝑒 to the power of five πœ‹ over four times 𝑖, all raised to the power of six. We can simplify this using the following relationship. Some number π‘₯ to the power of π‘Ž raised to the power of 𝑏 is equal to π‘₯ to the power of π‘Ž times 𝑏.

Using this relationship, we find that 𝑧 two to the power of six is equal to two to the power of two times six times 𝑒 to the power of five πœ‹ over four times six times 𝑖. Simplifying this further, we find that two times six is of course equal to 12. And five πœ‹ over four times six is equal to 15πœ‹ over two. Let’s replace these simplified values.

We can now perform the same process on 𝑧 three, which is raised to the power of negative four. As with the previous complex number, we can work on simplifying our powers. Here we have two to the power of three times negative four, which is of course equal to negative 12. And here in parentheses we have πœ‹ over four times negative four, which is equal to negative four πœ‹ over four or negative πœ‹. Let’s put in these simplified values.

We have now raised 𝑧 two and 𝑧 three to the relevant powers. But before we proceed, we’re going to perform one final simplification. You may notice that our complex numbers contain different fractional powers of 𝑒. Ignoring the multiple of 𝑖 leaves our four πœ‹ over three, 15πœ‹ over two, and negative πœ‹.

The next step of our calculation will be much easier if each of these are expressed in terms of a fraction with the same denominator. Since three, two, and one are all multiples of six, we’re gonna choose this as our denominator. Four πœ‹ over three is the same as eight πœ‹ over six. 15πœ‹ over two is the same as 45πœ‹ over six. And negative πœ‹ is equal to negative six πœ‹ over six.

Let’s rewrite each of our complex numbers, replacing the powers of 𝑒. Let’s now see why we’ve done this by moving on to the multiplication step. Since we have found 𝑧 one, 𝑧 two, and 𝑧 three to the relevant powers, we can input these into our equation. Here we have input our three complex numbers.

But we can reduce the calculation by remembering the following relationship. Some number π‘₯ raised to the power of π‘Ž multiplied by the same number π‘₯ raised to a different power 𝑏 is equal to π‘₯ to the power of π‘Ž plus 𝑏. This relationship allows us to form a sum for the power of two and the power of 𝑒. This is the same reason that we expressed our multiples all in terms of powers of two. And we change the fractional powers of 𝑒 all to have the same denominators.

Reducing the sums, we find the following answer. And we can now convert our two to the power of three back to an eight. Doing so, we find that 𝑧 one multiplied by 𝑧 two to the power of six divided by 𝑧 three to the power of four is equal to eight 𝑒 to the power of 47πœ‹ over six times 𝑖.

Although we have seemingly found the answer to our question, we do have one more step to perform. Here we can clearly see that our answer is in exponential form, with an π‘Ÿ-value of eight and a πœƒ-value of 47πœ‹ over six. Although we have a valid answer, generally, when working with complex numbers in exponential or in trigonometric form, we want our value of πœƒ to be greater than or equal to zero and less than two πœ‹.

By looking at this range in terms of sixths, we can see that our πœƒ-value is very clearly greater than 12πœ‹ over six. In order to fix this, we can use the following relationship, which takes advantage of the periodicity that occurs in the complex plane. In the complex plane, 𝑒 to the power of π‘–πœƒ is equal to 𝑒 to the power of 𝑖 times πœƒ plus two πœ‹π‘›, where 𝑛 is an integer value.

For our complex number, the value of πœƒ is 47πœ‹ over six. By choosing a negative value of 𝑛 β€” in this case, 𝑛 equals negative three β€” we can reduce our πœƒ-value to lie within the desired range. πœƒ plus two πœ‹ times negative three is equal to the following. Working through the calculation, we find the answer to this is 11πœ‹ over six, which does indeed lie within the desired range.

To recap, using this relationship for 𝑒, we have found that, in the complex plane, a πœƒ-value of 47πœ‹ over six is equivalent to a πœƒ-value of 11πœ‹ over six. This allows us to replace the πœƒ-value in our answer. Since πœƒ is now in the desired range, we have now reached the end of our journey. And we have found that 𝑧 one multiplied by 𝑧 two to the power of six divided by 𝑧 three to the power of four is equal to eight 𝑒 to the power of 11πœ‹ over six times 𝑖.

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