Lesson Video: Polar Form of Complex Numbers | Nagwa Lesson Video: Polar Form of Complex Numbers | Nagwa

Lesson Video: Polar Form of Complex Numbers Mathematics

In this video, we will learn how to represent a complex number in polar form, calculate the modulus and argument, and use this to change the form of a complex number.

17:50

Video Transcript

In this video, we’re going to learn about the polar form of a complex number. This is a way of writing a complex number which is particularly suited for problems involving multiplication. This new form is best understood using an Argand diagram. So let’s recap.

An Argand diagram represents all complex numbers on a plane. The real numbers lie on the horizontal axis or π‘₯-axis and the purely imaginary numbers lie on the vertical or 𝑦-axis. Which point on this diagram represents the complex number four minus four 𝑖? Well, we take the real part four to be the π‘₯-coordinate and the imaginary part negative four to be the 𝑦-coordinate. So the point representing four minus four 𝑖 is here in the fourth quadrant of the diagram.

We’ve also learnt about two properties of complex numbers: the modulus of a complex number and the argument of a complex number. Let’s start with the modulus. The modulus generalises the concept of absolute value of a number from the real numbers to the complex numbers. And it’s written in the same way with two vertical bars either side of the number. Writing a complex number in terms of its real part π‘Ž and imaginary part 𝑏, we have a formula for its modulus. This is the square root of π‘Ž squared plus 𝑏 squared. But really, this formula is best understood using the Argand diagram.

If we draw a vector from the origin to the point representing our complex number, then the modulus of our complex number is the length or magnitude of this vector and in the same way that the modulus of our complex number gives the magnitude of this vector. Its argument gives the direction of the vector. The argument of a complex number 𝑧 is written as arg 𝑧. And the formula for arg π‘Ž plus 𝑏𝑖 depends on the quadrant in which it lies, but in each case involves arctan of 𝑏 over π‘Ž. Again, the best way to think about the argument is using the Argand diagram.

It’s the angle of the vector measured counterclockwise from the positive real axis. So we start at the positive real axis and move counterclockwise until we get to our vector. This is the angle then. It’s not too hard to see that this acute angle has measure 45 degrees or πœ‹ by four radians. And by considering the angle at a point, our argument is 350 degrees or seven πœ‹ by four radians.

When talking about the argument of a complex number, we tend to use radians. And so, we can write that arg of four minus four 𝑖 is seven πœ‹ over four. We could also subtract two πœ‹ to get the principal argument negative πœ‹ by four, which you can think of as a measure of the orange acute angle with a minus sign because this angle needs to be measured in the opposite direction.

Having found the argument of our complex number, we might as well find its modulus. We find this using the formula we have. π‘Ž is four and 𝑏 is negative four. And simplifying, we get the square root of 32, which in simplest form is four root two. Let’s tidy up and interpret on our diagram. We found the modulus and argument of our complex number by using its real and imaginary parts. Well, actually, we got lucky with the argument because we noticed the 45-degree angle. But you can check that computing arctan of the imaginary part negative four over the real part four would have given the same answer.

A question we can ask is can we go the other way. Given the modulus and argument of a complex number 𝑧, can we find it’s real and imaginary parts and tend to write down our 𝑧? It feels like we ought to be able to. We know that the argument of our complex number is negative πœ‹ by four. And so, if we get our protractor out, we can see that our complex number must lie somewhere on this ray or line. And then, the modulus tells us how far along the line we need to go. We take our ruler and measure four root two units along from the origin to find that our complex number must lie here.

The idea that we can specify a point by giving its distance from the origin and direction that measured from the π‘₯-axis instead of its π‘₯- and 𝑦-coordinates gives rise to polar coordinates which you might know about. Applying the same idea on an Argand diagram to complex numbers gives rise to the polar form of a complex number. Let’s see how we can write a complex number 𝑧 in terms of its modulus and argument.

Given that the modulus of 𝑧 is π‘Ÿ and the argument of 𝑧 is πœƒ, find 𝑧.

We draw an Argand diagram to help us. And as the modulus of 𝑧 is π‘Ÿ, the point representing 𝑧 on the Argand diagram must be a distance of π‘Ÿ from the origin. So it lies on the circle with centre of the origin and radius π‘Ÿ. And as its argument is πœƒ, it must lie somewhere on this purple line too. So here is 𝑧 at the intersection. But have we really found 𝑧, what we should do is to write it down in the form π‘Ž plus 𝑏𝑖.

And to do that, we have to find the real part π‘Ž and the imaginary part 𝑏. We can read off the real part π‘Ž; it’s the π‘₯-coordinate of our point, and, similarly, for the imaginary part 𝑏. How do you write these π‘Ž and 𝑏 in terms of π‘Ÿ and πœƒ? Well, if you were in a unit circle, then π‘Ž would be cos πœƒ and 𝑏 would be sin πœƒ. But unfortunately, we’re not. The radius is π‘Ÿ. And so, everything is scaled up by π‘Ÿ, meaning that π‘Ž is π‘Ÿ cos πœƒ and 𝑏 is π‘Ÿ sin πœƒ.

We can see this in another way, noticing a right triangle with hypotenuse π‘Ÿ, the radius of the circle, side length π‘Ž adjacent to the angle πœƒ, and 𝑏 opposite it. Sin πœƒ is therefore the opposite 𝑏 over hypotenuse π‘Ÿ. And so, π‘Ÿ sin πœƒ equals 𝑏. And what we need to do is swap the sides. Similarly, cos πœƒ is the adjacent side length π‘Ž over the hypotenuse π‘Ÿ. And so, our cos πœƒ equals π‘Ž. Again, we just need to swap the sides to find that π‘Ž is π‘Ÿ cos πœƒ.

Having found the real and imaginary parts of 𝑧 in terms of π‘Ÿ and πœƒ, we can write 𝑧 in terms of π‘Ÿ and πœƒ just by substituting. 𝑧 is π‘Ÿ cos πœƒ plus π‘Ÿ sin πœƒ 𝑖. And this will do as the answer to our question. It means that the complex number 𝑧 whose modulus is π‘Ÿ and whose argument is πœƒ is represented by the point π‘Ÿ cos πœƒ, π‘Ÿ sin πœƒ on an Argand diagram. Those coordinates should look familiar if you’ve learnt about the polar coordinates. If we know the modulus and argument of a complex number, we can use this as a formula to find the complex number.

Given that the modulus of 𝑧 is four root two and the argument of 𝑧 is negative πœ‹ by four, find 𝑧.

Well, we see that the modulus π‘Ÿ is four root two and the argument πœƒ is negative πœ‹ by four. We substitute them into our formula. And we can simplify either by using a calculator or by using odd and even identities and special angles. Cos is an even function. And so, cos of negative πœ‹ by four is cos of πœ‹ by four. And πœ‹ by four is a special angle whose cosine we remember to be root two over two. Similarly, using the fact that sin is an odd function, we get that sin of negative πœ‹ by four is negative root two over two. Substituting these values and simplifying, we get four minus four 𝑖. But sometimes, we don’t want to simplify. We can rewrite our formula slightly by taking out the common factor of π‘Ÿ, getting π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ.

Notice that we’ve also swapped the order of 𝑖 and sin πœƒ here. And it turns out it’s very useful to write a complex number in this form. Applying this to our example, where the modulus of 𝑧 is four root two and its argument is negative πœ‹ by four, we write 𝑧 in the form π‘Ÿ times cos πœƒ plus 𝑖 times sin πœƒ. Just by substituting four root two for π‘Ÿ and negative πœ‹ by four for πœƒ, we get that 𝑧 is four root two times cos negative πœ‹ by four plus 𝑖 sin negative πœ‹ by four. This isn’t just a step of working on the way to writing down the value of 𝑧. It’s a valid way of expressing the value of 𝑧 in its own right.

Here is a definition then. When a complex number 𝑧 is written in the form π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ, it is said to be in β€œpolar form.” This can also be called β€œtrigonometric form” because it involves trigonometric functions cosine and sine or the β€œmodulus-argument form” as this form makes it so easy to read off the modulus and argument. With 𝑧 as in the definition, its modulus is π‘Ÿ and its argument is πœƒ.

This leaves us with the question of what’s the call of the original form π‘Ž plus 𝑏𝑖. This is called β€œalgebraic form,” β€œCartesian form,” or β€œrectangular form.” Four minus four 𝑖 is in this form. And four root two times cos negative πœ‹ by four plus 𝑖 sin negative πœ‹ by four is the same complex number written in polar form. Let’s now look at an example, where some numbers are written correctly in polar form and some numbers are not.

Which of the following complex numbers are correctly expressed in polar form?

Take a moment now to pause the video and look through each option carefully before we go through them together. Okay, are you ready? Here we go. Our complex number is said to be written in polar form if it’s written in the form π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ for some values of π‘Ÿ and πœƒ. We always want the value of π‘Ÿ to be greater than or equal to zero. So it is the modulus of our complex number. And sometimes, you want πœƒ to be in interval from negative πœ‹ to πœ‹ including πœ‹ but not negative πœ‹ so that it is the principal argument of the complex number. But we wouldn’t worry about that for the moment.

Okay, let’s start with option A; is this written in the required form? The value of root two outside the parentheses looks fine. But inside the parentheses, we have sin something plus 𝑖 cos something where we want cos something plus 𝑖 sin something. This is not correctly expressed in polar form.

How about option B? Well, we see that outside the parentheses the values of 𝑖 is five which is positive. That’s good news. And inside the parentheses, we have cos of something plus 𝑖 sin of something which is what we want. And importantly, in both cases, the somethings are the same; negative five πœ‹ by six is the value of πœƒ. Now, this value of πœƒ is negative. But that’s fine. That’s allowed. In fact, it’s even the principal argument of our complex number. And so, B is correctly expressed in polar form.

Moving on to C, we see that the value of π‘Ÿ is 𝑒 squared, again the positive number. But inside the parentheses, we have cos of something minus 𝑖 sin of something. And we would need that minus sin to be a plus for this complex number to be correctly expressed in polar form.

So we move on to D. The value of π‘Ÿ here is three πœ‹ by four. Inside the parentheses, we have cos of something plus 𝑖 sin of something as required. And again, those somethings are the same. The value of πœƒ is root 35. Now, you might think it’s strange that this complex number has a modulus of three πœ‹ by four and an argument of root 35. Surely, they should be the other way round, right? But technically speaking, there’s nothing wrong with this. This is in polar form. However, if you do ever write down a complex number like this where the modulus is a multiple of πœ‹ and the argument is the square root of a number, then you should probably make sure that you haven’t accidentally switch those two values.

Finally, option E, we have a positive value of π‘Ÿ outside the parentheses and cos of something plus 𝑖 sin of something inside the parentheses. But those somethings aren’t the same. And we have 35πœ‹ over seven and 35πœ‹ over six. These should be the same value πœƒ which is the argument of complex number. As they are not the same, this number is not correctly expressed in polar form.

Our answer is, therefore, that only B and D are correctly expressed in polar form. As an extension, you might like to use some identities you know about sine and cosine to correctly express options A and C in polar form. Unfortunately, there’s no easy way to do this for option E.

Let’s see a quick example of converting from a trigonometric form to rectangular form before we try to convert the other way.

Find cos πœ‹ by six. Find sin πœ‹ by six. And hence, express the complex number 10 cos πœ‹ by six plus 𝑖 sin πœ‹ by six in rectangular form.

Well, πœ‹ by six radians which is 30 degrees is a special angle. And so, we remember that cos of πœ‹ by six is root three by two and sin of πœ‹ by six is a half. Alternatively, your calculator might give you these values. And now that we have these two values, we can substitute them into our complex number in trigonometric form, getting 10 times root three over two plus a half 𝑖. And distributing that 10 over the terms in parentheses, we get five root three plus five 𝑖 which as required is in rectangular form, also known as algebraic form or Cartesian form, the form π‘Ž plus 𝑏𝑖. Now, let’s convert a complex number from algebraic form to polar form.

Find the modulus of the complex number one plus 𝑖. Find the argument of the complex number one plus 𝑖. And hence, write the complex number one plus 𝑖 in polar form.

We can draw an Argand diagram to help us. And we can draw the vector from the origin zero on the Argand diagram to the complex number one plus 𝑖. The modulus of one plus 𝑖 is just the magnitude of this vector. And by considering a right triangle and applying the Pythagorean theorem, we find that this is the square root of one squared plus one squared, which is the square root of two. Of course, the formula for the modulus of π‘Ž plus 𝑏𝑖 would have given us the same answer. That’s a modulus of one plus 𝑖. What about its argument?

Well, that’s a measure of this angle here, which we’ll call πœƒ. And because we have a right-angled triangle with opposite side length one and adjacent side length one, we know that tan πœƒ is equal to the opposite one of the adjacent one. And so, πœƒ is arctan one over one which is arctan one which is πœ‹ by four. We also could have seen this by noticing that we’re dealing with an isosceles right triangle. And so, πœƒ must be 45 degrees which is πœ‹ by four in radians. Now that we have the modulus and argument of our complex number, we can write it in polar form.

Our complex number is said to be written in polar form if it’s written in the form π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ. And importantly for us, if the complex number 𝑧 is written in this form, then its modulus is π‘Ÿ and its argument is πœƒ. Well, we know the modulus and argument of our complex number so we can just substitute these in as values of π‘Ÿ and πœƒ. The value of π‘Ÿ is the modulus root two and the value of πœƒ is the argument πœ‹ by four. This is a complex number one plus 𝑖 in polar form then. This is how we convert a number from algebraic form to polar form. We find its modulus and its argument and then we substitute these into the formula.

Let’s see another example.

Express the complex number 𝑧 equals four 𝑖 in trigonometric form.

We do this in three steps. We find π‘Ÿ which is the modulus of 𝑧. We find πœƒ which is its argument. And we substitute these values into 𝑧 equals π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ. But first, let’s draw an Argand diagram with four 𝑖 of course lying on the imaginary axis. We can see its modulus, its distance from the origin, is four. We could have got this using our formula instead. In any case, π‘Ÿ is four. How about its argument?

Trying to use a formula involving arctan 𝑏 over π‘Ž won’t work as π‘Ž is zero. And we can’t divide by zero. But luckily, we have our diagram where the argument is just this angle here, whose measure is 90 degrees or πœ‹ by two radians. So the argument of 𝑧 is πœ‹ by two. This is a value we have to substitute for πœƒ. And we are now ready to substitute. And doing so, we get four times cos πœ‹ by two plus 𝑖 sin πœ‹ by two.

Here are the key points that we’ve covered in the video. Much as points in the plane can be given using either Cartesian or polar coordinates, complex numbers can be given in either algebraic or polar form. The polar form of a complex number 𝑧 is 𝑧 equals π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ, where π‘Ÿ which is greater than or equal to zero is the modulus and πœƒ is the argument. To convert 𝑧 equals π‘Ž plus 𝑏𝑖 which is in algebraic form to polar form, we calculate its modulus π‘Ÿ and argument πœƒ and substitute these values into the above formula. To convert to 𝑧 equals π‘Ÿ times cos πœƒ plus 𝑖 sin πœƒ which is in trigonometric form to algebraic form, we evaluate the sine and cosine, distribute π‘Ÿ, and simplify.

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