Video: Operations with Complex Numbers in Polar Form

In this video, we will learn how to perform calculations with complex numbers in polar form.

17:54

Video Transcript

Previously, we learned about the polar form of a complex number. This is also known as the trigonometric or modulus argument form. We saw how to recognise when the complex number is written in this form and to how to convert between this form and the form π‘Ž plus 𝑏𝑖, known as the algebraic rectangular or Cartesian form of a complex number. But we haven’t seen why we should ever write a complex number in polar form. What makes it better than algebraic form? The answer or part of the answer is that polar form makes multiplication easy.

Let’s recap the polar form of a complex number. This is 𝑧 equals π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ, where π‘Ÿ must be greater than or equal to zero. With 𝑧 written in this form, we can just read off the modulus and argument. Its modulus is the value of π‘Ÿ. And its argument is the value of πœƒ. This explains why polar form is also known as modulus-argument form. This is best understood using an Argand diagram. The modulus π‘Ÿ is the distance at the point 𝑧 from the origin zero. And the argument πœƒ is the measure of the angle that the vector from zero to 𝑧 makes with the positive real axis, when measured counterclockwise. I’d like to show you how the polar form makes multiplication easy. So we’re going to need two complex numbers.

Let’s call this one 𝑧 one then instead of just 𝑧, with modulus π‘Ÿ one and argument πœƒ one, and introduce the second complex number, 𝑧 two, whose modulus is π‘Ÿ two and whose argument is πœƒ two, making it π‘Ÿ two times cos πœƒ two plus 𝑖 sin πœƒ two in polar form. Then, it turns out there’s a very simple expression for the product 𝑧 one 𝑧 two in polar form. This is π‘Ÿ one π‘Ÿ two times cos πœƒ one plus πœƒ two plus 𝑖 sin πœƒ one plus πœƒ two. This is in polar form, with a modulus of π‘Ÿ one times π‘Ÿ two. That’s the product of the moduluses of 𝑧 one and 𝑧 two. And its argument is πœƒ one plus πœƒ two, which is the sum of the arguments of 𝑧 one and 𝑧 two. So we multiply the moduluses but add the arguments. This gives a geometric interpretation to the product of complex numbers on an Argand diagram.

We can think of the effect on 𝑧 one of multiplying by 𝑧 two. We end up with a complex number 𝑧 one 𝑧 two, whose modulus is π‘Ÿ two times as big as 𝑧 ones and whose argument is πœƒ two more than 𝑧 ones. Multiplication by 𝑧 two therefore transforms the complex plane by dilation scale factor π‘Ÿ two, followed by rotation of πœƒ two counterclockwise. It’s beyond the scope of this video to discuss this any further. But you might like to think for yourself what happens when 𝑧 two is a positive real number or a negative real number or 𝑖 or an imaginary number.

We’ve made this claim about the multiplication of complex numbers in polar form. And now we should prove it. We do this by first using the expressions for 𝑧 one and 𝑧 two in polar form. We can reorder the factors so that π‘Ÿ two appears next to π‘Ÿ one. And now we just have to worry about the terms in parentheses. And this is just the multiplication of two complex numbers in the algebraic form. We get cos πœƒ one cos πœƒ two plus 𝑖 times cos πœƒ one sin πœƒ two, notice how we’ve reordered the factors so that 𝑖 is at the front, plus 𝑖 sin πœƒ one cos πœƒ two plus 𝑖 squared sin πœƒ one sin πœƒ two. And 𝑖 squared is negative one. So this should be minus sin πœƒ one sin πœƒ two at the end. Using this fact, we can group the real terms and imaginary terms.

And now, for the main part of the proof, we recognise the real and imaginary parts from the trigonometric angle sum identities. The real part is exactly cos πœƒ one plus πœƒ two. And the imaginary part after swapping the terms is just sin of πœƒ one plus πœƒ two. And this is what we had to prove. These trigonometric identities seem to come from out of the blue to give us an unexpectedly nice answer. We get 𝑧 one times 𝑧 two in polar form. And we can read off the modulus and argument. The modulus is π‘Ÿ one times π‘Ÿ two. And the argument is πœƒ one plus πœƒ two.

Now, if we look back at our claim, we can see that π‘Ÿ one was just the modulus of 𝑧 one and π‘Ÿ two was the modulus of 𝑧 two. So the modulus of 𝑧 one times 𝑧 two is the modulus of 𝑧 one times the modulus of 𝑧 two. The modulus of the product is the product of the moduluses. And similarly, πœƒ one was the argument of 𝑧 one and πœƒ two of 𝑧 two. So the argument of 𝑧 one times 𝑧 two is the argument of 𝑧 one plus the argument of 𝑧 two. And knowing the modulus and argument of this product allows us to write it in polar form. So these two facts taken together are equivalent to the claim that we proved. And now that we have proved this claim, let’s apply it.

Given that 𝑧 one equals two times cos πœ‹ by six plus 𝑖 sin πœ‹ by six and 𝑧 two equals one over root three times cos πœ‹ by three plus 𝑖 sin πœ‹ by three, find 𝑧 one times 𝑧 two.

If 𝑧 one equals π‘Ÿ one times cos πœƒ one plus 𝑖 sin πœƒ one and 𝑧 two equals π‘Ÿ two times cos πœƒ two plus 𝑖 sin πœƒ two, then 𝑧 one times 𝑧 two is π‘Ÿ one times π‘Ÿ two times cos πœƒ one plus πœƒ two plus 𝑖 sin πœƒ one plus πœƒ two. Comparing what we have with our formula then, we see that π‘Ÿ one is two. And π‘Ÿ two is one over root three. And the modulus of our product, π‘Ÿ one times π‘Ÿ two, is therefore two times one over root three. How about the argument? Well, we can see that πœƒ one is πœ‹ by six. And πœƒ two is πœ‹ by three. Our argument is their sum, πœ‹ by six plus πœ‹ by three. And now, we just need to simplify. Two times one over root three is two over root three. And we can rationalise this denominator if we want to by multiplying both numerator and denominator by root three, to get two root three over three. How about the argument? We can write πœ‹ by three as two πœ‹ by six. And hence, the argument is three πœ‹ by six or πœ‹ by two.

Making the substitution, we see that the product we’re looking for is two root three over three times cos πœ‹ by two plus 𝑖 sin πœ‹ by two. We’ll leave our answer in polar form even though we might know the values of cos πœ‹ by two and sin πœ‹ by two. And hopefully, having gone through this example, you will agree that multiplying two numbers in polar form is very easy. You just have to multiply their moduluses and add their arguments. That’s less work than multiplying two numbers in the algebraic form. Let’s see another example.

If 𝑧 one equals seven times cos πœƒ one plus 𝑖 sin πœƒ one, 𝑧 two equals 16 times cos πœƒ two plus 𝑖 sin πœƒ two, and πœƒ one plus πœƒ two equals πœ‹, then what is 𝑧 one times 𝑧 two?

Well, we know that the modulus of 𝑧 one times 𝑧 two will be the modulus of 𝑧 one, that’s seven, times the modulus of 𝑧 two. That’s 16. And seven times 16 is 112. And we also know that the argument of 𝑧 one times 𝑧 two is the argument of 𝑧 one plus the argument of 𝑧 two. The argument of 𝑧 one is πœƒ one. And the argument of 𝑧 two is πœƒ two. So this becomes πœƒ one plus πœƒ two, which we’re told in the question is πœ‹. Now we know the modulus and argument of this product. It’s easy to write it down in polar form. If the modulus of 𝑧 is π‘Ÿ and the argument of 𝑧 is πœƒ, then we can write 𝑧 in polar form as π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ. In our case, π‘Ÿ is 112. And πœƒ is πœ‹. Substituting then, we find that 𝑧 one times 𝑧 two is 112 times cos πœ‹ plus 𝑖 sin πœ‹. This is our answer in polar form.

We could also write this answer as negative 112. That will be acceptable too. And we could’ve found this value either by using the fact that cos πœ‹ is negative one and sin πœ‹ is zero or by using the fact that if a complex number has an argument of πœ‹, that means it’s a negative real number. And of course, the only negative real number with modulus 112 is negative 112.

We’ve seen that using the polar form of the complex number makes multiplication very easy. And the same is true for division. If 𝑧 one equals π‘Ÿ one times cos πœƒ one plus 𝑖 sin πœƒ one and 𝑧 two equals π‘Ÿ two times cos πœƒ two plus 𝑖 sin πœƒ two, then 𝑧 one over 𝑧 two, the quotient of 𝑧 one and 𝑧 two, is π‘Ÿ one over π‘Ÿ two times cos πœƒ one minus πœƒ two plus 𝑖 sin πœƒ one minus πœƒ two. Equivalently, the modulus of a quotient of complex numbers is the quotient of their moduluses. And the argument of a quotient of complex numbers is the difference of their arguments. And it’s this equivalent statement that we’ll prove.

To prove this, we’ll use what we know about the modulus and argument of a product of complex numbers. Given two complex numbers, 𝑀 one and 𝑀 two, we know that the modulus of their product is the product of their moduluses. And the argument of their product is the sum of their arguments. This we’ve proved already. The key idea of this proof is to let 𝑀 one equal 𝑧 one over 𝑧 two and 𝑀 two equal 𝑧 two. Making this substitution, we find that the modulus of 𝑧 one over 𝑧 two times 𝑧 two is the modulus of 𝑧 one over 𝑧 two times the modulus of 𝑧 two. And on the left-hand side, the 𝑧 twos cancel. And so we get just the modulus of 𝑧 one on the left-hand side. And now, we can rearrange this to make the modulus of 𝑧 one over 𝑧 two the subject. And when we do this, we find that it is the modulus of 𝑧 one over the modulus of 𝑧 two, as required.

Now, we just need to do the same for the argument. Substituting 𝑧 one over 𝑧 two for 𝑀 one and 𝑧 two for 𝑀 two, we get that the argument of 𝑧 one over 𝑧 two times 𝑧 two is the argument of 𝑧 one over 𝑧 two plus the argument of 𝑧 two. And again, the 𝑧 twos on the left-hand side cancel. So we get just the argument of 𝑧 one on the left-hand side. Rearranging, we find that the argument of 𝑧 one over 𝑧 two is the argument of 𝑧 one minus the argument of 𝑧 two as required. We’ve managed to prove this equivalent statement, finding the modulus and argument of our quotient. And with this modulus and argument, we can write down the polar form of the quotient 𝑧 one over 𝑧 two. This can also be proved like fully multiplication case just by writing the quotient of the two complex numbers in polar form. This is more straightforward but involves a lot of algebraic manipulation. Anyway, now that we’ve seen how we can divide complex numbers in polar form, let’s apply it to some problems.

Given that 𝑧 one equals 20 times cos πœ‹ by two plus 𝑖 sin πœ‹ by two and 𝑧 two equals four times cos πœ‹ by six plus 𝑖 sin πœ‹ by six, find 𝑧 one over 𝑧 two in polar form.

Well, we know in general that the modulus of a quotient of complex numbers is the quotient of the moduluses. And we can just read off the moduluses. They are 20 and four. So the of modulus of our quotient is 20 divided by four, which is five. And we can use the fact that the argument of a quotient is the difference of the arguments, which in our case are πœ‹ by two and πœ‹ by six. And we can write πœ‹ by two over the denominator six. It’s three πœ‹ by six. So the difference is two πœ‹ by six, which is πœ‹ by three. That’s the argument of our quotient. Now that we have the modulus and argument, we can write our quotient in polar form. 𝑧 one over 𝑧 two is five times cos πœ‹ by three plus 𝑖 sin πœ‹ by three. Now, isn’t this much easier than dividing complex numbers in algebraic form, where you have to multiply both numerator and denominator by the complex conjugate of the denominator? I think so.

Given that 𝑧 equals cos seven πœ‹ over six plus 𝑖 sin seven πœ‹ over six, find one over 𝑧.

Now, we could divide in the way that we divide a complex number in algebraic form, using the complex conjugate of that number to make the denominator real. But 𝑧 is also in polar form. Its modulus one is not explicitly written. But we can write it in. And we can see that the argument is seven πœ‹ by six. Now, if you write one in polar form, we’ll have a quotient of two complex numbers in polar form, which we know how to evaluate. The modulus of one is one. And the argument of one and indeed any positive real number is zero. So one in polar form is one times cos zero plus 𝑖 sin zero.

Now, we have to find the reciprocal of 𝑧. That’s one over 𝑧. And we can write both one and 𝑧 in polar form. The modulus of their quotient is then one divided by one, which is one. And their argument is the difference of the arguments. That’s zero minus seven πœ‹ by six, which is negative seven πœ‹ by six. We don’t need to write the modulus of one explicitly. We can just write that one over 𝑧 is cos of negative seven πœ‹ over six plus 𝑖 sin of negative seven πœ‹ over six. This is our final answer.

We’ve seen that using the polar form of complex numbers makes multiplication and division much easier. They don’t require as much work. However, we haven’t seen how to add or subtract complex numbers in polar form. This is because addition and subtraction of complex numbers in polar form is in general much harder than of complex numbers in algebraic form. To add two complex numbers in algebraic form, you just add the real parts and the imaginary parts and similarly for subtraction. It couldn’t get much easier.

If you do have to add or subtract complex numbers in polar form, it’s generally best to convert them to algebraic form, before adding or subtracting and then converting back to polar form if required. Before the video ends, we’re going to see another operation which polar form makes easier. This is exponentiation, which involves finding powers of a complex number. This is a bit of a sneak peek of a topic which will be covered in more depth later.

Consider the complex number 𝑧 equals one plus 𝑖 root three.

Our first task is to find the modulus of 𝑧. That’s straightforward. We know that the modulus of π‘Ž plus 𝑏𝑖 is the square root of π‘Ž squared plus 𝑏 squared. Substituting one for π‘Ž and root three for 𝑏, we get the square root of one plus three, which is the square root of four, which is two. Now, for the argument of 𝑧, we see that 𝑧 is in the first quadrant. And so its argument πœƒ can be found using arctan. Substituting again, we find that the argument is arctan root three over one, which a calculator will tell us is πœ‹ by three. Hence, that means using the previous results, use the properties of multiplication of complex numbers in polar form to find the modulus and the argument of 𝑧 cubed.

Let’s not rush straight to 𝑧 cubed but first consider 𝑧 squared. 𝑧 squared is just 𝑧 times 𝑧. And we know that the modulus of the product of complex numbers is the product of their moduluses. So we find that the modulus of the square of 𝑧 is the modulus of 𝑧 squared. How about the argument of 𝑧 times 𝑧? This is the argument of 𝑧 plus the argument of 𝑧 or two times the argument of 𝑧. Now, we’re ready to consider 𝑧 cubed. We use the fact that 𝑧 cubed is 𝑧 times 𝑧 squared. And so the modulus of 𝑧 cubed is the modulus of 𝑧 times the modulus of 𝑧 squared, which we know to be the modulus of 𝑧 squared. And so we see that the modulus of the cube of a complex number is the cube of the modulus of that complex number. To find the argument of 𝑧 cubed, we use the same trick, writing 𝑧 cubed as 𝑧 times 𝑧 squared, using what we know about the argument of a product and the argument of 𝑧 squared, finding that the argument of the cube of a complex number is three times the argument of that complex number.

Now, we just need to substitute the known values of the modulus of 𝑧 and the argument of 𝑧. The modulus of 𝑧 is two. So the modulus of 𝑧 cubed is eight. And the argument of 𝑧 is πœ‹ by three. So the argument of 𝑧 cubed is πœ‹. Finally, we find the value of 𝑧 cubed. We know the modulus and argument of 𝑧 cubed. So we can just write 𝑧 cubed down in polar form. It’s eight times cos πœ‹ plus 𝑖 sin πœ‹. Or using the fact that cos πœ‹ is negative one and sin πœ‹ is zero, we could write it as negative eight. More interesting than the final answer is what we found on the way to it, those being the expressions for the modulus and argument of 𝑧 cubed. This leads to a very compact expression for the cube of the complex number in polar form. This is a special case of de Moivre’s theorem, which will be explored in more detail at another time.

The key points that we’ve covered in this video are as follows. Calculations involving multiplication and division of complex numbers are often simpler when we work in polar form. However, for addition and subtraction, this is definitely not true. For complex numbers, 𝑧 one equals π‘Ÿ one times cos πœƒ one plus 𝑖 sin πœƒ one and 𝑧 two equals π‘Ÿ two times cos πœƒ two plus 𝑖 sin πœƒ two, the following rules hold. Their product, 𝑧 one 𝑧 two, is π‘Ÿ one times π‘Ÿ two times cos πœƒ one plus πœƒ two plus 𝑖 sin πœƒ one plus πœƒ two. And their quotient, 𝑧 one over 𝑧 two, is π‘Ÿ one over π‘Ÿ two times cos πœƒ one minus πœƒ two plus 𝑖 sin πœƒ one minus πœƒ two. From these formulas, we can read off the modulus and argument of a product and quotient of complex numbers. The modulus of the product of complex numbers is the product of their moduluses. And the argument of the product of complex numbers is the sum of their arguments. The modulus of a quotient of complex numbers is the quotient of their moduluses. And the argument of the quotient is the difference of the arguments.

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