[Complex Numbers] Ex 1E Q22) | [Solved] (1 Viewer)

A1La5 · Sep 2, 2021

Hello, I am stuck on this question.

I feel like I am on the right track with this, but I'm missing something small. My working out is as follows (sorry for it being all over the place..) - what have I missed?

Drdusk · Sep 3, 2021

A1La5 said:
Hello, I am stuck on this question.

View attachment 31910

I feel like I am on the right track with this, but I'm missing something small. My working out is as follows (sorry for it being all over the place..) - what have I missed?

View attachment 31911

https://boredofstudies.org/threads/harder-complex-number-question.181881/

Exact same question was answered here.

CM_Tutor · Sep 3, 2021

@A1La5, $\arg{(\omega_1 - \omega_2)}$ is not the exterior angle of the triangle that you have drawn. Leaving aside that this argument is negative and the angle that you are calling the exterior angle is positive, the exterior angle of a triangle at a vertex must be between one of the sides that meet at that vertex and the other side the meets at that vertex produced.

The theorem you need is that co-interior angles on parallel lines are supplementary, with the angles being

$\phi = -\arg{(\omega_1 - \omega_2)}$ (as $\phi$ is a positive angle in a triangle but $\omega_1 - \omega_2$ has a negative argument),

$\alpha_2$ , and the third angle in the triangle that you have not named. Let's call it $\beta$ , so that your triangle has angles $\alpha_2 - \alpha_1$ , $\theta$ , and $\beta$ (though you don't actually need $\theta$ ).

You first need to find $\beta$ , so that a second application of co-interior angles with $\phi$ , $\beta$ , and $\alpha_2$ allows you to find $\phi$ .

As you have recognised that $\omega_1 - \omega_2$ is a vector from $\omega_2$ to $\omega_1$ , you need only complete the parallelogram with $\omega_1 + \omega_2$ as the second diagonal (i.e., with vertices at $O$ , $\omega_1$ , $\omega_2$ , and $\omega_1 + \omega_1$ ).

Note that the interior angles of this parallelogram are $\alpha_2 - \alpha_1$ and $2\beta$ , and these angles are supplementary, which gives you that

$\beta = \cfrac{\pi + \alpha_1 - \alpha_2}{2}$

and from that you can find the size of

$\phi = \pi - \beta - \alpha_2 = \cfrac{2\pi - (\pi + \alpha_1 - \alpha_2) - 2\alpha_2}{2} = \cfrac{\pi - \alpha_1 - \alpha_2}{2}$

and hence you can determine the principal argument of $\omega_1 - \omega_2$ .

CM_Tutor · Sep 3, 2021

It's worth noting that the thread that @Drdusk points to takes a different approach, finding my angle $\beta$ without completing the parallelogram by instead using the given fact that $|\omega_1| = |\omega_2|$ . This makes the triangle that @A1La5 drew above an isosceles triangle, from which it follows that $\beta = \theta$ as these are its base angles. Both approaches are valid, of course.

A1La5 · Sep 3, 2021

CM_Tutor said:
@A1La5, $\arg{(\omega_1 - \omega_2)}$ is not the exterior angle of the triangle that you have drawn. Leaving aside that this argument is negative and the angle that you are calling the exterior angle is positive, the exterior angle of a triangle at a vertex must be between one of the sides that meet at that vertex and the other side the meets at that vertex produced.

The theorem you need is that co-interior angles on parallel lines are supplementary, with the angles being

$\phi = -\arg{(\omega_1 - \omega_2)}$ (as $\phi$ is a positive angle in a triangle but $\omega_1 - \omega_2$ has a negative argument),

$\alpha_2$ , and the third angle in the triangle that you have not named. Let's call it $\beta$ , so that your triangle has angles $\alpha_2 - \alpha_1$ , $\theta$ , and $\beta$ (though you don't actually need $\theta$ ).

You first need to find $\beta$ , so that a second application of co-interior angles with $\phi$ , $\beta$ , and $\alpha_2$ allows you to find $\phi$ .

As you have recognised that $\omega_1 - \omega_2$ is a vector from $\omega_2$ to $\omega_1$ , you need only complete the parallelogram with $\omega_1 + \omega_2$ as the second diagonal (i.e., with vertices at $O$ , $\omega_1$ , $\omega_2$ , and $\omega_1 + \omega_1$ ).

Note that the interior angles of this parallelogram are $\alpha_2 - \alpha_1$ and $2\beta$ , and these angles are supplementary, which gives you that

$\beta = \cfrac{\pi + \alpha_1 - \alpha_2}{2}$

and from that you can find the size of

$\phi = \pi - \beta - \alpha_2 = \cfrac{2\pi - (\pi + \alpha_1 - \alpha_2) - 2\alpha_2}{2} = \cfrac{\pi - \alpha_1 - \alpha_2}{2}$

and hence you can determine the principal argument of $\omega_1 - \omega_2$ .

Damm, looks like I was a lot further from the solution then I thought. Thanks so much for your help!

EDIT: @CM_Tutor, would the horizontal line drawn from $\omega_2$ in the diagram be parallel to the positive real axis? If so, then can we also use alternate angles + the angle sum of a triangle (with angles $\phi$ , $\theta$ , and $\alpha_2$ ) to calculate the value of $\phi$ ?

CM_Tutor · Sep 3, 2021

A1La5 said:
EDIT: @CM_Tutor, would the horizontal line drawn from $\omega_2$ in the diagram be parallel to the positive real axis?

Yes, absolutely. It is the fact that it is parallel to the $x$ -axis that allows me to say that $\alpha_2$ and $\phi + \beta$ are supplementary co-interior angles. This gets you the result that

$\phi + \beta + \alpha_2 = \pi$

from which you can find $\phi$ once an expression is found for $\beta$ in terms of $\alpha_1$ , $\alpha_2$ , and $\pi$ .

If so, then can we also use alternate angles + the angle sum of a triangle (with angles $\phi$ , $\theta$ , and $\alpha_2$ ) to calculate the value of $\phi$ ?

I am not seeing what pair of parallel lines you mean. In your second diagram, it looks like the angle you have labelled as $\arg{(\omega_1 - \omega_2)}$ (that is, the angle that I have called $\phi$ ) and your angle $\theta$ might be alternate angles on parallel lines... but looks can be deceiving. If you look at your first diagram, with the circle, it is clear that the two lines are definitely not parallel as the first one is parallel to the real axis (as you noted above) and the second is inclined at angle $\alpha_1$ to the real axis.

However, if you were to produce the side of the triangle that is represented by $\omega_1 - \omega_2$ to the real axis to form a new triangle with one side on the real axis, then yes you could use alternate angles to establish that $\phi$ is one of the angles of the new triangle, but the others would be $\alpha_2$ and $\beta$ and the angle $\theta$ would not be an angle of the new triangle. (Well... except for the fact that the original triangle is isosceles and hence $\beta = \theta$ .)

This would be an alternative way to establish that $\phi + \beta + \alpha_2 = \pi$ , but you would still be left needing to find an expression for $\beta$ in terms of $\alpha_1$ , $\alpha_2$ , and $\pi$ to ultimately find $\phi$ and hence the required principal argument.

A1La5 · Sep 3, 2021

CM_Tutor said:
I am not seeing what pair of parallel lines you mean.

Yeah, guess I should've clarified. The two parallel lines that I'm talking about would be the horizontal line drawn from $\omega_2$ and the horizontal real axis. I redrew an extra set of diagrams to help you decipher what I mean.

In this diagram I have marked the two parallel sides I am referring to.

First, I used the fact that the triangle in the original diagram is isosceles (as both $\omega_1$ and $\omega_2$ have modulus 1) to express $\theta$ (or $\beta$ ) in terms of $\pi$ , $\alpha_1$ and $\alpha_2$ . This is:

$\theta= \cfrac{\pi + \alpha_1 - \alpha_2}{2}$

Now, I redrew a triangle from this above diagram below.

The side opposite $\theta$ in this triangle would be the side on the real axis you brought up in your response. Using alternate angles the third angle here must be your angle $\phi$ , and then you use the angle sum of this triangle to express $\phi$ in terms of $\pi$ and $\theta$ , and therefore in terms of $\alpha_1$ and $\alpha_2$ as we know how to express $\theta$ in terms of $\alpha_1$ and $\alpha_2$

After this the result just falls out, using your definition of $\phi = -\arg{(\omega_1 - \omega_2)}$ . Hopefully there shouldn't be any holes in my reasoning.

CM_Tutor · Sep 3, 2021

Sounds solid to me, @A1La5 .

[Complex Numbers] Ex 1E Q22) | [Solved] (1 Viewer)

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