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# FP2 locus of complex numbers: watch

1. Suppose we had an equation arg(z-3/z)=pi/4.

The textbook and examsolutions require sketching and stuff but I'm terrible sketching (which is why I take the long method in solving mod inequalities by playing around with the definition and using a very long method) but for this method, I encountered a new method that is easier (For me) but I'm not too sure as whether or not this would get the full marks as I'm not sure if it'd be on the alternative solutions.

arg(z-3 / z) = pi/4 --> arg(z-3) - arg = pi/4 --> arg[(x-3)+iy] - arg(x+iy) = pi/4

Taking tans of both sides and using tan(A-B) = tan(A)-tan(B) / 1+tan(A)(B) yields y/x-3 - y/x / 1 + y/x-3 . y/x = 1 --> x^2+y^2-3x-3y = 0 --> (x-3/2)^2 + (y-3/2)^2 = 9/2 e.g a circle radius 3/root 2 and centre (3/2,3/2) but z-3/z needs to have both real and imaginary parts positive to have an argument of pi/4 hence z-3/z = (x-3)+iy / x+iy --> x(x-3)+y^2 > 0 e.g. (x-3/2)^2 + y^2 > 9/4.

The final solution to the question is the locus is the part of the circle (3/2,3/2) radius 3/root 2 which lies outside the other circle centre (3/2,0) and radius 3/2 (this can help sketching it)

Does anybody have a link to learn a much better, quicker method than the textbook explanation an this?
2. (Original post by thekidwhogames)
Suppose we had an equation arg(z-3/z)=pi/4.

The textbook and examsolutions require sketching and stuff but I'm terrible sketching (which is why I take the long method in solving mod inequalities by playing around with the definition and using a very long method) but for this method, I encountered a new method that is easier (For me) but I'm not too sure as whether or not this would get the full marks as I'm not sure if it'd be on the alternative solutions.

arg(z-3 / z) = pi/4 --> arg(z-3) - arg = pi/4 --> arg[(x-3)+iy] - arg(x+iy) = pi/4

Taking tans of both sides and using tan(A-B) = tan(A)-tan(B) / 1+tan(A)(B) yields y/x-3 - y/x / 1 + y/x-3 . y/x = 1 --> x^2+y^2-3x-3y = 0 --> (x-3/2)^2 + (y-3/2)^2 = 9/2 e.g a circle radius 3/root 2 and centre (3/2,3/2) but z-3/z needs to have both real and imaginary parts positive to have an argument of pi/4 hence z-3/z = (x-3)+iy / x+iy --> x(x-3)+y^2 > 0 e.g. (x-3/2)^2 + y^2 > 9/4.

The final solution to the question is the locus is the part of the circle (3/2,3/2) radius 3/root 2 which lies outside the other circle centre (3/2,0) and radius 3/2 (this can help sketching it)
Seems like a fine approach. What approach does the book suggest?

To make your approach the best it can be:
You can skip the tan compound angle by just some quick and accurate manipulation to get into the form

so

Argument of this is which implies hence .

We know that our values lie on this circle.
Then just sketch that on the complex plane. And as you say, we require and . So, hence , as well as so

These conditions yield the loci, which are marked orange here:
Spoiler:
Show

3. (Original post by RDKGames)
Seems like a fine approach. What approach does the book suggest?

To make your approach the best it can be:
You can skip the tan compound angle by just some quick and accurate manipulation to get into the form

so

Argument of this is which implies hence .

We know that our values lie on this circle.
Then just sketch that on the complex plane. And as you say, we require and . So, hence , as well as so

These conditions yield the loci, which are marked orange here:
Spoiler:
Show

I think the book does it by considering half lines and playing around with circle theorems to get the loci and equation. Thanks for the tips! Would you recommend using this method in FP2 problems for 4/5 marks because it's easy to do but longer?
4. (Original post by thekidwhogames)
I think the book does it by considering half lines and playing around with circle theorems to get the loci and equation. Thanks for the tips! Would you recommend using this method in FP2 problems for 4/5 marks because it's easy to do but longer?
If you can confidently carry it out with no mistakes, sure.

Though it's useful to get to know the half-line + circle theorem approaches.
5. (Original post by RDKGames)
If you can confidently carry it out with no mistakes, sure.

Though it's useful to get to know the half-line + circle theorem approaches.
Alright thanks for the help!

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