### Video Transcript

Given that π§ is equal to negative five plus nine π, find the principal argument of π§ rounded to the nearest two decimal places.

In this question, weβre given a complex number π§, given in algebraic form, and weβre asked to find the principal argument of our complex number π§. And we need to round our answer to two decimal places.

To answer this question, letβs start by recalling what we mean by an argument of a complex number π§ and what it means for this argument to be the principal argument. First, we say the argument of a complex number π§ is equal to π when itβs the angle that π§ makes with the positive real axis on an Argand diagram.

And thereβs a few things worth pointing out about this definition. First, when we say the angle that π§ makes with the positive real axis on an Argand diagram, really what we mean is the angle that a ray from the origin to π§ will make with the positive real axis. However, it can be easier to think about this as just the angle that π§ makes.

Next, when we measure this angle, we measure it counterclockwise to be positive and clockwise to be negative. And just like with any angle measured in this way, thereβs going to be multiple angles which give the same value. So to get around this, we introduce something called the principal argument of a complex number. We say that π is the principal argument of our complex number π§ if when measured in radians π is bigger than negative π and less than or equal to π or when measured in degrees π is bigger than negative 180 degrees and less than or equal to 180 degrees.

We can now use this information to find the principal argument for our complex number π§. Weβre going to start by sketching this onto an Argand diagram. Remember, in an Argand diagram, the horizontal axis represents the real part of our complex number and the vertical axis represents the imaginary part of our complex number. We want to plot the complex number π§ is equal to negative five plus nine π onto our Argand diagram.

So we need to find the real and imaginary parts of this number. And we can do this because π§ is given in algebraic form. Thatβs the form π plus ππ, where π and π are real numbers. The real part of π§ will be our value of π. Thatβs the constant on its own, in this case negative five. And the imaginary part of π§ will be our coefficient of π, which in this case is nine. We can use this to plot π§ onto our Argand diagram. Its horizontal coordinate should be negative five, and its vertical coordinate should be nine.

Now, weβre going to want to sketch the principal argument of π§ onto our diagram. To do this, we start by connecting π§ to the origin with a ray. Remember, the argument of an angle π§ is the argument this ray makes with the positive real axis. The angle we want to find, π, is as shown. And we can see since this is measured counterclockwise, this value of π will be positive.

Thereβs actually a lot of different methods we can use to find this value of π. All of these will involve some kind of trigonometry. The easiest way is to do this directly from our diagram. Weβre going to find the value of the angle πΌ. We can find the angle πΌ by constructing the following right-angled triangle. The height of this right-angled triangle is the modulus of the imaginary part of π§, which is nine. And the width of this right-angled triangle is the modulus of the real part of π§. Thatβs five. So in this right-angled triangle, we know the opposite length to angle πΌ and the adjacent length to angle πΌ. And we know, by using trigonometry, the tan of πΌ is the length of its opposite side divided by the length of its adjacent side in a right-angled triangle.

So in our case, the tan of πΌ is equal to nine over five. And we can find the value of πΌ by taking the inverse tangent of both sides. πΌ is equal to the inverse tan of nine over five. And weβll find this value in degrees. We get that πΌ is equal to 60.945 and this continues degrees. Finally, we can find the value of π by looking at our diagram. We know that angle πΌ plus angle π makes up a straight line. This is going to be equal to 180 degrees. Therefore, π is going to be 180 degrees minus πΌ, which we can calculate is 119.054 and this continues degrees.

And the question wants us to give our answer to two decimal places. So to do this, we look at the third decimal place in our expansion, which is four, to determine if we need to round up or round down. Since four is less than five, we need to round down. And this gives us our final answer. Therefore, we were able to find the principal argument of π§ is equal to negative five plus nine π rounded to two decimal places. We got 119.05 degrees.