# Question Video: Finding the Argument of Complex Numbers in Terms of Pi Mathematics

What is the argument of the complex number ππ, where π < 0?

03:24

### Video Transcript

What is the argument of the complex number ππ, where π is less than zero.

In this question, weβre asked to find the argument of a complex number. And weβre given the form of this complex number. Itβs in the form ππ, where our value of π is negative. To answer this question, weβre first going need to recall what we mean by the argument of a complex number. First, we recall the argument of a complex number π§, written arg π§, is the angle that π§ makes with the positive real axis on an Argand diagram. And there are a few things worth pointing out about this definition.

First, technically, itβs not the angle that π§ makes with the positive real axis on our Argand diagram. Itβs actually the angle that the ray from the origin to π§ makes with this axis. However, it can be useful to think of it in this way. Next, we usually measure this angle in radians. However, we can also measure it in degrees if we prefer. And when our angle is positive, this means we measured it counterclockwise. And if we give a negative angle, then we measured our angle clockwise.

Next, much like with any other angle measured in this manner, thereβs going to be several different angles which represent the same value. For example, a turn of zero, a turn of two π, and a turn of four π will all represent the same angle. So usually we choose our argument to be greater than negative π and less than or equal to π.

So to find the argument of the complex number given to us in the question, weβre first going to need to sketch an Argand diagram. Remember, in an Argand diagram, our horizontal axis represents the real part of our complex number and the vertical axis represents the imaginary part of our complex number. In our case, we want to plot the number ππ onto our Argand diagram. We can see that ππ is an imaginary number. It has no real component. So the real part of ππ is going to be equal zero. Similarly, the imaginary part of ππ is going to be the coefficient of π, which in this case is π. And itβs worth pointing out here we know that our value of π is negative.

Remember in an Argand diagram, the horizontal coordinate represents the real part of our imaginary number and the vertical coordinate represents the imaginary part of our complex number. So for our value of ππ whose real part is zero and imaginary part is π, its coordinates on our Argand diagram is going to be zero, π, where of course our value of π is negative. So we need to draw this on the negative part of our imaginary axis.

We want to find the argument of this complex number, so it can help to add the line segment from the origin to our value of ππ on our Argand diagram. Then the argument of π§ is going to be the angle between this line segment and the positive real axis. And weβll measure this clockwise because it makes the measure of our angle the smallest. Normally, with complex numbers, we would need to use trigonometry to help us find the argument. However, itβs not necessary in this case. We can see that this is just a right angle.

And in radians, a right angle is given by π by two. But remember, weβre measuring this angle clockwise, so this needs to be negative π by two. And this gives us our final answer. The argument of ππ is equal to negative π by two. Now we could stop here, but weβve actually proven a very useful result. Weβve proven if π is less than zero, then the argument of ππ will be equal to negative π by two. So we can actually use this result to evaluate the argument of any complex number given to us in this form. We wouldnβt need to sketch this onto an Argand diagram and then find the resulting angle. Although, this is still a good idea anyway.

Therefore, in this question, we were able to prove a result about finding the argument of certain complex numbers. We were able to show the argument of any complex number in the form ππ, where our value of π is less than zero, is negative π by two.