Video: Finding the Argument of a Complex Number in Radians

Find the argument of the complex number 2 βˆ’ 7𝑖 in radians. Give your answer correct to two decimal places.

03:08

Video Transcript

Find the argument of the complex number two minus seven 𝑖 in radians. Give your answer correct to two decimal places.

We have been given a complex number in rectangular or algebraic form. In general, we can say a complex number is in this form if it is π‘Ž plus 𝑏𝑖. And we’re being asked to find its argument. We’ll begin by considering what the number two minus seven 𝑖 looks like on the Argand diagram.

Remember, this is a way of representing complex numbers graphically. We have the horizontal axis, which represents the real component of our number. And the vertical axis represents its imaginary part. We can see then that the number two minus 𝑧 [seven] 𝑖 must lie in the fourth quadrant. And if we join this point to the origin, the argument is the angle this line makes with the horizontal.

In fact, we measure this in a counterclockwise direction. So this tells us the value of our argument, let’s call that πœƒ, is going to be negative. And we know it’s negative rather than a large value of πœ‹ because we generally represent our complex numbers using the principal argument. That’s πœƒ is greater than negative πœ‹ and less than or equal to πœ‹.

Let’s now add a right-angle triangle. We can see that the side adjacent to the included angle πœƒ must be two units. And the side opposite from the included angle is seven units. And since this is a right-angle triangle for which we know the measure of two of its sides and we’re looking to find the missing angle, we can use right-angle trigonometry.

Here, tan πœƒ is equal to opposite over adjacent. So we can say that, for our value of πœƒ, tan πœƒ is equal to seven over two. And we solve this equation for πœƒ by finding the inverse tan of both sides of the equation. The inverse tan of tan πœƒ or the arctan of tan πœƒ is πœƒ. So we can see that πœƒ is equal to the inverse tan of seven over two. And as long as we have our calculator working in radians, we get πœƒ to be equal to 1.2924 and so on. And we said that our value of πœƒ needs to be negative. So πœƒ is negative 1.29 radians.

Now, this is a rather long-winded method. And in fact, we can generalize it for the complex number of the form π‘Ž plus 𝑏𝑖. We say that, for this complex number, its argument is the arctan or inverse tan of 𝑏 divided by π‘Ž. Let’s see what that looks like with our complex number.

The constant π‘Ž or the real part is two. And the coefficient of 𝑖 or the imaginary component is negative seven. 𝑏 is negative seven. So in this case, we say that πœƒ is equal to inverse tan of negative seven over two. And if we type that into our calculator, we get negative 1.29 radians. It’s much more sensible to use this formula when finding the argument of the complex number. But we can use an Argand diagram to check our answer.

In this case, πœƒ or the argument of our complex number is negative 1.29 radians.

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