Complex Locus (1 Viewer)

[khorne]

arg(z-i) - arg(z+i) = pi/2

Using algebra, I get x^2 + y^2 = 1, y=/= 1 or -1

But according to the answers it's y=/= 1, -1 and x > 0 (i.e a semicircle)...I don't get the x> 0 part...

 

[gurmies]

The necessary condition should be x < 0.

 

[khorne]

I thought x should be unrestricated, since, at the end of the algebra,

arctan((-2x)/(x^2 + y^2 1 1)) = pi/2, i.e x^2 + y^2 = 1

but since arg((z-i)/(z+i)) cannot be zero/underfined, z =/= i or - i


Btw small mistake, fixed it in original q.

 

[gurmies]

Technically x < 0 is still a true statement. For the other branch of the circle, where x > 0, the governing equation would be arg(z-i)-arg(z+i)=-pi/2 or, with some rearranging, arg(z+i)-arg(z-i)=pi/2.

 

[khorne]

Ah right...I guess this is the problem with doing it algebraically =[

 

[gurmies]

Also as an aside, if you tan both sides of the equation, you find yourself dealing with tan[pi/2], which is of course, undefined.

 

[khorne]

but if it's undefined, we can state x^2 + y^2 -1 = 0

 

[gurmies]

I'm trying to think why, algebraically, the condition is true. There is definitely an explanation.

 

[cyl123]

arctan((-2x)/(x^2 + y^2 -1)) = pi/2
==> denominator=0 AND numerator is positive as tan(pi/2) is +infinity
So -2x>0 and x^2+y^2=1
So x<0 and x^2+y^2=1

 

[gurmies]

Thank you cyl123, clever.

 

[jet]

arctan((-2x)/(x^2 + y^2 -1)) = pi/2
==> denominator=0 AND numerator is positive as tan(pi/2) is +infinity
So -2x>0 and x^2+y^2=1
So x<0 and x^2+y^2=1

Just to clarify, technically it isn't mathematically 'rigorous' to say that denominator = 0, since that is mathematically not allowed (for want of a better phrase). You should formally write it as 'as the denominator approaches 0, arctan( insert fraction here) --> pi/2'.

 

[cyl123]

I agree. A more rigorous solution would be:

arg(z-i) - arg(z+i) = pi/2
let z=x+iy , x,y are real
arg((z-i)/(z+i))=pi/2
arg((x+iy-i)/(x+iy+i)=pi/2
arg((x^2+y^2-1-2ix)/denominator (dont care))=pi/2
==> real part =0 and imaginary part >0
we also note the denominator (after making it real) will always be positive as it is the sum of 2 squares
So we end up with x^2+y^2-1=0 and -2x>0
x^2+y^2=1 and x<0

Personally, I would avoid the arctan stuff and solve geometrically

 

[khorne]

Thanks for the help so far, but I have one more question...

I've attached the worked solutions for that question, and while I get the geometric approach, I am concerned about the vector? z-i. On the diagram, they have labelled the z-i vector's argument at the x-axis. However, isn't z+i the vector from i to z, thus the argument should be a negative number between the top of z+i and the parallel of the x-axis? Or otherwise, are they even vectors?

Otherwise, how could you do this geometrically? I tried following the method layed out in Cambridge, but to no avail (I get the semicircle on the other side, not the same one)

My geometric working:

Let z+i (a) and z-i (b) meet at p.

construct z-i at a such that it is parallel to the original. If it's on the x+ side, the diagram shows z+i - pi/2, but if it's on the -ve, it shows z+i + pi/2.

 

[jet]

z + i is the vector from z to -i. To check, just plug in -i and you get 0, which means -i is the point from which you are drawing the vector.

EDIT: The solution looks perfectly fine to me.

EDIT2: In general, any vector z - w is the vector FROM w TO z. A simple proof just plugs w where z is (as z is any complex number) and you get 0. You interpret this as saying that the vector (z - w) has no length when it's 'head' is at the point w, which must mean that it's tail is also at w.

 

[khorne]

Yes, so that is also true for z-i, as +i is where i'm drawing it from, but this, however, means that if you extended z+i, to the a-axis the possitive direction is not infact the argument, and so their working out is wrong?

 

[jet]

Nah, because if you think about it, the vector from -i TO z points to the top-right, meaning that it has a positive argument.

EDIT: Though, they are wrong in the argument for the vector z - i, since that points to the bottom right, making the argument negative. They should have 180 - phi = arg(z - i), if you're going by that diagram.

 

[khorne]

I know, I was refering to the vector z-i, not z+i, i.e i to z.

 

[jet]

Lol, read my edit above.

EDIT again: If you think about it, flipping it across the y-axis would change the signs of the arguments of the vectors, meaning that they would be true, thus confirming the algebraic explanation.

My reasoning is such: Using the diagram they have drawn, the real argument of the vector z - i should be 180 - phi., let's call this gamma. But, we assume that gamma is positive, so arg (z - i) = -gamma. Then, by the angle sum of that triangle, theta - gamma + 90 = 180, which implies that arg(z + i) - arg(z - i) = pi/2. By flipping the diagram across the y-axis, the signs of the arguments should change, and we get arg(z - i) - arg( + i) = pi/2, which is the original statement.

 

[cyl123]

The vector z-a is the vector from the complex number a to the complex number z IN THE DIRECTION TOWARDS a. The problem with the solutions is that they mistaken the direction of the vector. z-i is going FROM i TO P, while the argument they have shown is for the vector going from P to i, which gives i-z. So the solutions are wrong as I see it.

For these sort of questions, there is a nice method to it. The main part of these questions is to identify which side of the line joining i and -i is the arc on. So to do so easily, pick a point on one side of the line where it is easy to draw vectors from to ensure the angle at P is above the x axis so we are dealing with positive angles only. It doesn't matter where you choose it as long as it is on one side. For this example, choose P on the diagram I drawn (locus1).

In the diagram, ensure the vectors are pointing TOWARDS the chosen point P. Now we note theta represents arg(z+i) (or arg(z-(-i))) and phi represents arg(z-i). The angle at P should represent the angle stated in the equation (in this case pi/2). We should have arg(z-i)-arg(z+i)=pi/2 but according to the position of P, we have it the wrong way around. Thus we know the locus will be on the other side of the interval joining i and -i (More specifically, the y axis).

Note be sure to rearrange the formula so the angle is positive

I shall do another example to illustrate the idea.

Consider Arg(z+i)-Arg(z-1+2i)=-pi/4
First rearrange the formula so the angle is positive:
Arg(z-(1-2i))-Arg(z+i)=pi/4
Now we note that the locus will be on one side of the interval joining 1-2i and -i (labelled I on the diagram). So choose a point on one side where it is easy to draw vectors TOWARDS that point such that the angle at the point will be positive. From the diagram (locus2), I will choose a point on the right hand side at P.

(refer to locus2) theta represents Arg(z-1+2i) while phi represents Arg(z+i). the angle at P should be pi/4. From the equation, we should have theta-phi=pi/4, which holds true on the diagram. (Note the value of the angle does not matter at the moment as long as it is positive. We are only concerned with which side of the interval the locus is on. Now we know the locus is on the right hand side of the interval, thus drawing the rest accounting for the size of the angle is easy. Full version on locus3.

I hope you understand

NB sorry for the crappy paint drawings

 

[khorne]

Thanks very much guys...I feel pretty foolish for not getting that...Rep for all