# Loci in the complex plane

• Mar 15th 2009, 01:59 PM
Erghhh
Loci 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 PM
Prove It
Quote:

Originally Posted by Erghhh
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?

$z = x + iy$

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

Upon realising the denominator, this gives

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

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

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

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

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

Evaluate this and set it equal to $45^{\circ}$. This gives you the locus of the circle. You should then be able to find it's radius.
• Mar 15th 2009, 07:59 PM
mr fantastic
Quote:

Originally Posted by Erghhh
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?

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 $\sqrt{2}$ 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 AM
mr fantastic
Quote:

Originally Posted by Prove It
$z = x + iy$

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

Upon realising the denominator, this gives

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

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

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

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

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

Evaluate this and set it equal to $45^{\circ}$. This gives you the locus of the circle. You should then be able to find it's radius.

This will give the circle upon which the locus lies. But the locus itelf is only a subset of this circle.

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

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

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

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

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

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

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 $z = -1$, 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 = $\frac{| -1 - 1|}{2 \sin \frac{\pi}{4}} = \sqrt{2}$.

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

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