Video Transcript
Given that the modulus of 𝑧 is 𝑟 and the argument of 𝑧 is 𝜃, find 𝑧.
We draw an Argand diagram to help us. And as the modulus of 𝑧 is 𝑟, the point representing 𝑧 on the Argand diagram must be a distance of 𝑟 from the origin. So it lies on the circle with centre of the origin and radius 𝑟. And as its argument is 𝜃, 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 𝑎 plus 𝑏𝑖.
And to do that, we have to find the real part 𝑎 and the imaginary part 𝑏. We can read off the real part 𝑎; it’s the 𝑥-coordinate of our point, and, similarly, for the imaginary part 𝑏. How do you write these 𝑎 and 𝑏 in terms of 𝑟 and 𝜃? Well, if you were in a unit circle, then 𝑎 would be cos 𝜃 and 𝑏 would be sin 𝜃. But unfortunately, we’re not. The radius is 𝑟. And so, everything is scaled up by 𝑟, meaning that 𝑎 is 𝑟 cos 𝜃 and 𝑏 is 𝑟 sin 𝜃.
We can see this in another way, noticing a right triangle with hypotenuse 𝑟, the radius of the circle, side length 𝑎 adjacent to the angle 𝜃, and 𝑏 opposite it. Sin 𝜃 is therefore the opposite 𝑏 over hypotenuse 𝑟. And so, 𝑟 sin 𝜃 equals 𝑏. And what we need to do is swap the sides. Similarly, cos 𝜃 is the adjacent side length 𝑎 over the hypotenuse 𝑟. And so, our cos 𝜃 equals 𝑎. Again, we just need to swap the sides to find that 𝑎 is 𝑟 cos 𝜃.
Having found the real and imaginary parts of 𝑧 in terms of 𝑟 and 𝜃, we can write 𝑧 in terms of 𝑟 and 𝜃 just by substituting. 𝑧 is 𝑟 cos 𝜃 plus 𝑟 sin 𝜃 𝑖. And this will do as the answer to our question. It means that the complex number 𝑧 whose modulus is 𝑟 and whose argument is 𝜃 is represented by the point 𝑟 cos 𝜃, 𝑟 sin 𝜃 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 by four, find 𝑧.
Well, we see that the modulus 𝑟 is four root two and the argument 𝜃 is negative 𝜋 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 by four is cos of 𝜋 by four. And 𝜋 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 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 𝑟, getting 𝑟 times cos 𝜃 plus 𝑖 sin 𝜃.
Notice that we’ve also swapped the order of 𝑖 and sin 𝜃 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 by four, we write 𝑧 in the form 𝑟 times cos 𝜃 plus 𝑖 times sin 𝜃. Just by substituting four root two for 𝑟 and negative 𝜋 by four for 𝜃, we get that 𝑧 is four root two times cos negative 𝜋 by four plus 𝑖 sin negative 𝜋 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.
