# Complex function : composition of a translation, rotation, etc

• September 12th 2009, 12:42 PM
arbolis
Complex function : composition of a translation, rotation, etc
I don't know how to approach the following problem : Write the transformation $T(z)=\frac{i-z}{i+z}$ as a composition of translations, rotations, dilatations and inversions. Use the result to find the image of the semi-plane $y>0$.

My attempt : My idea was to write $T(z)$ in the form $re^{i \pi \theta}$ but I didn't reach anything important I believe.
So I guess there's a better way to approach the problem. I'd like to know it/them.
• September 12th 2009, 05:18 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
I don't know how to approach the following problem : Write the transformation $T(z)=\frac{i-z}{i+z}$ as a composition of translations, rotations, dilatations and inversions. Use the result to find the image of the semi-plane $y>0$.

My attempt : My idea was to write $T(z)$ in the form $re^{i \pi \theta}$ but I didn't reach anything important I believe.
So I guess there's a better way to approach the problem. I'd like to know it/them.

If you are interested in conformal mapping you can read the entire detailed thread here.

$\frac{i-z}{i+z} = -\frac{z-i}{z+i} = - \frac{z+i-2i}{z+i} = -1 + \frac{2i}{z+i}$

Let $f: z\mapsto z+i$, let $g:z\mapsto \tfrac{1}{z}$, let $h: z \mapsto i$, let $F: z\mapsto 2z$, let $G:z\mapsto z-1$. Thus, $f$ is translation, $g$ is inversion, $h$ is rotation (by $\tfrac{\pi}{2}$), $F$ is dilation, and $G$ is translation.

However, $G\circ H \circ h\circ g \circ f: z\mapsto \tfrac{i-z}{i+z}$.
• September 12th 2009, 05:40 PM
arbolis
Thanks a lot TPH! (I'll take some time to read through the link... seems really interesting. I don't know if my course will introduce me conformal mapping, it is made for physicists.)
Just a little question... why did you bother to rewrite $\frac{i-z}{i+z}$ in the form $-1 + \frac{2i}{z+i}$?
I see that you got rid of a z over z form... is it a general way to solve problem like these?

Quote:

Originally Posted by ThePerfectHacker
If you are interested in conformal mapping you can read the entire detailed thread here.

$\frac{i-z}{i+z} = -\frac{z-i}{z+i} = - \frac{z+i-2i}{z+i} = -1 + \frac{2i}{z+i}$

Let $f: z\mapsto z+i$, let $g:z\mapsto \tfrac{1}{z}$, let $h: z \mapsto i$, let $F: z\mapsto 2z$, let $G:z\mapsto z-1$. Thus, $f$ is translation, $g$ is inversion, $h$ is rotation (by $\tfrac{\pi}{2}$), $F$ is dilation, and $G$ is translation.

However, $G\circ H \circ h\circ g \circ f: z\mapsto \tfrac{i-z}{i+z}$.

• September 12th 2009, 06:04 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
I see that you got rid of a z over z form... is it a general way to solve problem like these?

Yes, we have to clear the z in the numerator. Because the basic inversion is 1/z, it has no z in numerator. Thus, we have to bring it into that form.
• September 12th 2009, 06:23 PM
arbolis
Quote:

Originally Posted by ThePerfectHacker
Yes, we have to clear the z in the numerator. Because the basic inversion is 1/z, it has no z in numerator. Thus, we have to bring it into that form.

Thanks. I'm infinitely thankful to you.
• September 12th 2009, 06:38 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
Thanks. I'm infinitely thankful to you.

You can bow before my feet and say, "Oh Master how much I love thee". (Nod)
• September 12th 2009, 06:46 PM
arbolis
Quote:

Originally Posted by ThePerfectHacker
You can bow before my feet and say, "Oh Master how much I love thee". (Nod)

(Ninja)
(Bow)