# Visual Complex Analysis

1. Jun 20, 2005

### Thomas2054

"Visual Complex Analysis"

I have gotten myself wound around the axel regarding something in "Visual Complex Analysis" (Dr. Tristan Needham) that should be easy.

On p. 18 (paperback edition), towards the bottom, the result for two rotations about different points has got me stumped. I cannot see my way clear to processing the two transfer functions in succession. Basically I am struggling with what to multiply and what to add in what order so as to end up with a rotation about the origin followed by a translation.

Suggestions would be appreciated?

Thanks.

Best regards,

Thomas

2. Jun 20, 2005

### mathwonk

lets ee here, a rotation is an isometry of the plane as is a translation. indeed every orientation preserving isometry is a composition of these two: proof: given an orientation preserving isometry f such that f(0) = a, compose f with translation from a to 0, and get an orientation preserving iusometry taking 0 to 0, which must of course (who could doubt it) a rotation.

i'm tired of this game. does this help?

3. Jun 20, 2005

### Thomas2054

I think I understand the concept, it is the mechanics of the calculation that I am stuck on.

Do you by chance have a copy of the book? If not, I can e-mail a scan of the page or put it on my web site so you can look at the equation I am have trouble with, if you are willing to spend a few minutes.

When Needham does a single rotation about a non-zero complex point, he shows it done as a translation to the origin, a rotation about the origin, and then a translation back to the original point.

Then he shows how this can be done as a rotation along with the translation point "adjusted" for not being done at the origin.

OK, I am good to that point and I can reproduce the calculations, i.e., I understand what he has done.

When he does two rotations, about different points, I cannot do the math. I am an engineer and I am stuck on standard operations. I think I have an internal assumption and I am blocked from seeing a straightforward approach.

Thomas

4. Jul 18, 2011

### JungleJesus

Re: "Visual Complex Analysis"

I have the book as a pdf. The process of composing rotations is conceptually simple. Recall that a single rotation about an arbitrary point is a translation to the origin, a rotation about the origin, and then the inverse of the translation to the origin. Composing two rotations means you simply calculate the sequence of translations and rotations that you get from expanding the two rotations.

If Ra, th is a rotation of th about the complex number a, and Rb, sg is a rotation of sg about b, then each rotation is expressible as:
Ra, th = Ta o Rth, 0 o T-a and
Rb, sg = Tb o Rsg, 0 o T-b
where Ta is just a translation by a.

If I expand a rotation into an equation, I get:
Ra, th = a + ei*th(z - a)

Notice how this works: we decompose the sequence of translations and rotations from right to left (that's what the notation really means. Think of the rightmost entity as the deepest layer of parenthesis). The rightmost motion is the first entity to act on the number z, so z shows up there in the calculation. Every motion to the left then acts on the previous motion, giving a sequence of motions which is reflected in the formula above.

If Ra, th is first and Rb, sg is second, then the sequence of basic motions (translations and origin rotations) is:
(Tb o Rsg, 0 o T-b) o (Ta o Rth, 0 o T-a) = Tb o Rsg, 0 o T-b o Ta o Rth, 0 o T-a

Now simply expand this into an equation:
Rb, sg o Ra, th = b + ei*sg((-b) + a + ei*th(z-a))

If you distribute the multiplications, you will get a rotation of z by th + sg about the origin plus a constant (a rotation followed by a translation). Of course, with a little algebra, this can be expressed as a single rotation about a constant point (the trick is, however, that th + sg cannot be an integer multiple of 2*pi since that would yield a rotation of infinite radius, which is just a movement about a strait line, a pure translation.

Hope this helped!