Transformation

**Simplicity** · June 9th 2008, 11:49 AM

The transformation $T$ from the $z$ -plane, where $z = x + iy$ , to the w-plane, where...

$w=\frac{z+i}{z}, \ z \ne 0$

(a) The transformation $T$ maps the points on the line with equation $y=x$ in the $z$ -plane, other than $(0, 0)$ , to points on a line $l$ in the $w$ -plane. Find a cartesian equation of $l$ .

(b) Show that the image, under $T$ , of the line with equation $x+y+1=0$ in the $z$ -plane is a circle $C$ in the $w$ -plane, where $C$ has cartesian equation $u^2+v^2-u+v=0$ .

**bobak** · June 9th 2008, 01:37 PM

Originally Posted by Air

The transformation $T$ from the $z$ -plane, where $z = x + iy$ , to the w-plane, where...

$w=\frac{z+i}{z}, \ z \ne 0$

(a) The transformation $T$ maps the points on the line with equation $y=x$ in the $z$ -plane, other than $(0, 0)$ , to points on a line $l$ in the $w$ -plane. Find a cartesian equation of $l$ .

The complex number $z$ has an argument of $\frac{ \pi}{4}$ for $\Im (z) > 0$ and $- \frac{ 3 \pi }{4}$ for $\Im (z) < 0$ .

Consider $w - 1$ which is $\frac{i}{z}$ therefore $\arg(w -1) = \arg \left( \frac{i}{z} \right)$

$\Rightarrow \arg(w -1) = \arg(i) - \arg(z)$
$\Rightarrow \arg(w -1) = \frac{\pi}{2} - \arg(z)$

for $\Im (z) > 0$ $\arg(w -1) = \frac{\pi}{2} - \frac{\pi}{4} \ \ \Rightarrow \ \ \arg(w -1) = \frac{\pi}{4}$

for $\Im (z) < 0$ $\arg(w -1) = \frac{\pi}{2} + \frac{3 \pi}{4} \ \ \Rightarrow \ \ \arg(w -1) = \frac{5 \pi}{4}$

You should be able to pull out an equation form that.

Bobak

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