Question Video: Finding the Argument of a Complex Number in Terms of Pi | Nagwa Question Video: Finding the Argument of a Complex Number in Terms of Pi | Nagwa

Question Video: Finding the Argument of a Complex Number in Terms of Pi Mathematics • Third Year of Secondary School

What is the argument of the complex number −6?

03:14

Video Transcript

What is the argument of the complex number negative six?

In this question, we’re asked to find the argument of a complex number. And that complex number is negative six. And there’s something worth pointing out here. We might not be used to thinking of negative six as a complex number. However, any real number can be thought of as a complex number with imaginary part equal to zero. So any question we can ask about complex numbers we can also ask about real numbers.

In this question, we want to find the argument of this complex number. So let’s start by recalling what we mean by the argument of a complex number. We recall the argument of some complex number 𝑧, written arg of 𝑧, is said to be equal to some value of 𝜃 if 𝜃 is the angle that 𝑧 makes with the positive real axis on an Argand diagram. And there’s a few things worth pointing out about this definition.

First, really, the argument of a complex number is the angle that the ray connecting the origin to our point 𝑧 on our Argand diagram makes with the positive real axis. Next, we measure this angle counterclockwise to be positive and clockwise to be negative. And usually we measure this angle in radians. However, we can also measure this angle in degrees. And the last thing worth pointing out is the argument of a complex number 𝑧 is not unique. Just like with any other angle we measure in this manner, there are lots of equivalent angles. For example, the angle zero, 360 degrees, and 720 degrees are all equivalent.

So to find the argument of a complex number, we’re going to start by wanting to plot this onto an Argand diagram. Remember, in an Argand diagram, the horizontal axis represents the real parts of our complex number and the vertical axis represents the imaginary parts of our complex number. Since we want to find the argument of negative six, which is a real number, we already know these values. The real part of negative six is equal to itself, negative six. And since this is a real number, it has no imaginary part. The imaginary part of negative six is zero.

So on our point, we want the imaginary part to be equal to zero and the real part to be equal to negative six. So this just lies on the real axis at the point negative six. And if we want, we can give this point a name. We can call our complex number 𝑧. But this is not strictly necessary. Now, to find the arguments of this complex number, we’re going to need to connect it to the origin via a ray or line segment. And the argument of our complex number is going to be any angle that this line segment makes with the positive real axis. And it doesn’t really matter which one we choose. They’ll all be the arguments of our complex number.

We’ll choose to mark our angle counterclockwise and make it as small as possible. Usually, to find the argument of a complex number, we use trigonometry. However, it’s not necessary in this case. We can see that 𝜃 is just the angle in a straight line. And because this is measured counterclockwise, we know it will be positive. So 𝜃 is just going to be equal to 𝜋.

Therefore, we have shown the argument of the complex number negative six is equal to 𝜋. However, it’s worth pointing something out here. This would be true for any negative real number. For example, if we wanted to find the argument of a real number 𝑎 which was negative, then we can construct our line segment in exactly the same way and choose the exact same angle 𝜃, which means we’ve actually proven a result. The argument of any negative real number is equal to 𝜋. So we could’ve just used this result to answer our question directly. Therefore, we were able to show two different ways of finding the argument of the complex number negative six. In both cases, we were able to show this was given by 𝜋.

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