# [Complex plane] arg[(z-1)/(z+1)] = pi/3

1. Mar 29, 2014

### tusher

1. The problem statement, all variables and given/known data
Show that arg[(z-1)/(z+1)] represents a circle. Find it's radius and centre.

2. Relevant equations

3. The attempt at a solution

using z = (x+iy) i narrowed down to (z-1)/(z+1) = (iy)/(1+x) , assuming it was a circle.
What next?
Is this correct approach??

2. Mar 29, 2014

### Simon Bridge

The next step is to take the argument and show that it is the same as the equation for a circle.

3. Mar 29, 2014

### tusher

I miscalculated. :(

The steps are..

$\frac{z-1}{z+1}$ = $\frac{(x+iy-1)}{(x+iy+1)}$
= $\frac{(x+iy-1)(x-iy-1)}{(x+iy+1)(x-iy-1)}$
=$\frac{(x-1)^{2}-(iy)^{2}}{(x^{2})-(iy)^{2}}$
= $\frac{(x)^{2}-2x+1+y^{2}}{x^{2}-1-2iy-(iy)^{2}}$
=$\frac{x^{2}+y^{2}-2x+1}{x^{2}+y^{2}-2iy-1}$

Supposing,x$^{2}$ + y$^{2}$ = 1
we get ,
(x-1)/iy

I have no idea what to do next.

4. Mar 29, 2014

### Simon Bridge

Here, let me tidy that up a bit: \begin{align} C &=\frac{z-1}{z+1} & & \text{(1)}\\ &= \frac{(x+iy-1)}{(x+iy+1)} & & \text(2)\\ &= \frac{(x+iy-1)(x-iy-1)}{(x+iy+1)(x-iy-1)} & & \text(3)\\ &=\frac{(x-1)^{2}-(iy)^{2}}{(x^{2})-(iy)^{2}} & & \text(4)\\ &= \frac{(x)^{2}-2x+1+y^{2}}{x^{2}-1-2iy-(iy)^{2}} & & \text(5)\\ & =\frac{x^{2}+y^{2}-2x+1}{x^{2}+y^{2}-2iy-1} & & \text(6) \end{align}
(... I've called the original function C and numbered the lines to make it easier to talk about.)

Note: at some stage you'll have to find arg[C] and show that it is a circle.

... how does that follow? Again, I don't see your reasoning.
(Assume that I know the maths but I'm in another country on the other side of the World and we may have different conventions in how we approach math problems here. I won't be offended.)

...lets see: $$c=\frac{x-1}{iy}=-i\frac{x-1}{y}$$... is purely imaginary so the argument is: $\text{Arg}[c]=\pm\frac{\pi}{2}$ ... i.e. it is a couple of points, not a circle.

Perhaps you should end up with something like: $\text{Arg}[c] = x^2+y^2-k$ ... where $r=\text{Arg}[c]+k$ is the (real) radius?

It's kinda hard to see how that would work.
How can an argument come to a circle - it takes 2D and turns it into 1D?
So there is something missing from the problem statement.

Generally: $$\frac{z-z_1}{z-z_2}=c$$... is a circle in the complex plane if c ≠ 1 and is real.

... overall you need to think how the argument comes into this.
Note: C=r is the equation of a circle, radius r, with it's center at (x,y)=(1,0).
If A=arg[C] then the radius of the circle is r=√tan(A/2).
... depending on how the argument is defined in your course.

Aside: good use of the equation editor.
If you hot "quote" under this post you get to see how I tidied that up for you ;)

5. May 5, 2017

### Samik

So if we consider A(-1,0) and B(1,0) to be 2 points on the argand plane and z be any arbitrary point then line segment zA will be z+1 and zB will be z-1. The arguments will be arg(z+1) and arg(z-1) respectively. So the angle between the 2 lines will be arg((z-1)/z+1)). We have been given that this angle is constant(π/2). We know that any angle subtended by the end points of a chord on a circle is always equal. So we know that locus of z is a circle.

The center can be found out be taking the incenter of an equilateral triangle(one of the many possibilities of triangle zAB)whose base is AB and radius will be the distance from this point to any of the 2 points(A or B)

$$\frac{z-1}{z+1} = \frac{x^2+y^2-1 + i 2y}{x^2+y^2+2x+1}.$$