Given that arg((z-1)/(z+1))=45deg.

find the centre and radius of the circle?

i equated z-1=z+1, but i'm pretty sure that was the wrong method as i didn't get the equation of a circle.

Starting point, anyone?

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- Mar 15th 2009, 01:59 PMErghhhLoci in the complex plane
Given that arg((z-1)/(z+1))=45deg.

find the centre and radius of the circle?

i equated z-1=z+1, but i'm pretty sure that was the wrong method as i didn't get the equation of a circle.

Starting point, anyone? - Mar 15th 2009, 03:42 PMProve It
$\displaystyle z = x + iy$

So $\displaystyle \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$

Upon realising the denominator, this gives

$\displaystyle \frac{[(x - 1) + iy][( x+ 1) - iy]}{(x + 1)^2 + y^2}$

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

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

$\displaystyle = \frac{x^2 + y^2 - 1}{(x + 1)^2 + y^2} + i\frac{2y}{(x + 1)^2 + y^2}$

Remember that $\displaystyle \arg{Z} = \arctan{\frac{\mathbf{I}(Z)}{\mathbf{R}(z)}}$.

Evaluate this and set it equal to $\displaystyle 45^{\circ}$. This gives you the locus of the circle. You should then be able to find it's radius. - Mar 15th 2009, 07:59 PMmr fantastic
At the moment I only have time to say the following:

This is not a circle. It's the major arc of a circle.

The arc joins z = 1 and z = -1 (these points are NOT included in the locus) and lies on a circle of radius $\displaystyle \sqrt{2}$ and centre located on the perpendicular bisector of the chord joining z = 1 and z = -1 at a distance of 1 from this chord. - Mar 16th 2009, 02:43 AMmr fantastic
This will give the circle upon which the locus lies. But the locus itelf is only a subset of this circle.

In fact, since $\displaystyle \frac{z - 1}{z + 1}$ cannot have an argument of $\displaystyle \frac{\pi}{4}$ unless its real and imaginary parts are BOTH positive, the solution has to satisfy the following two implied restrictions:

$\displaystyle Re \left[ \frac{z - 1}{z + 1} \right] > 0 \Rightarrow x^2 + y^2 > 1$ .... (1)

$\displaystyle Im \left[ \frac{z - 1}{z + 1} \right] > 0 \Rightarrow y > 0$ .... (2)

The locus is therefore the major arc of the circle I have previously described and which can be found by solving $\displaystyle Re \left[ \frac{z - 1}{z + 1} \right] = Im \left[ \frac{z - 1}{z + 1} \right]$. As well as this algebraic approach suggested by**Prove It**, the parameters of this circle can also be found from geometric considerations:

$\displaystyle arg(z - 1) - arg(z + 1) = \frac{\pi}{4}$ is the locus of the intersection point P of the rays $\displaystyle arg(z - 1) = \alpha$ and $\displaystyle arg(z + 1) = \beta$ under the restriction $\displaystyle \alpha - \beta = \frac{\pi}{4}$.

It's not hard to show that these two rays only intersect when $\displaystyle 0 < \beta < \frac{3 \pi}{4} $, in which case the angle between these two rays is always $\displaystyle \frac{\pi}{4}$.

Using the the well known theorem that all angles subtended by the same chord of a circle at the circumference are equal, it follows that the locus is the major arc joining but not including z = -1 and z = 1. Using simple trigonometry on the right triangle*AOC*(where a is the point representing $\displaystyle z = -1$,*O*is the origin and*C*is the centre of the circle) it is easily seen that the arc lies on a circle with the following parameters:

1. Radius = $\displaystyle \frac{| -1 - 1|}{2 \sin \frac{\pi}{4}} = \sqrt{2}$.

2. Centre C located on the Imaginary axis at a distance $\displaystyle \frac{| -1 - 1|}{2 \tan \frac{\pi}{4}} = 1$ from the origin.

Several other approaches to finding the locus are possible, including a vector approach (using the dot product) and a parametric method.