complex analysis: linear transformations

Printable View

February 16th 2009, 11:50 AM
srw899

complex analysis: linear transformations

A function of the form f(z)=az+b where a and b are complex constants is called a linear transformation. show that every linear transformation can be expressed as the composition of a magnification, a ratation, and a translation. deduce from this that a linear transformation maps lines to lines and circles to circles.

Hint: write a in polar form.
February 16th 2009, 01:01 PM
manjohn12

Quote:

Originally Posted by srw899

A function of the form f(z)=az+b where a and b are complex constants is called a linear transformation. show that every linear transformation can be expressed as the composition of a magnification, a ratation, and a translation. deduce from this that a linear transformation maps lines to lines and circles to circles.

Hint: write a in polar form.

Let $a = re^{i \theta}$ and $z = be^{i \gamma}$ . Then $az = rbe^{i(\theta + \gamma)}$ . And let $b = b_{1}+ i b_2$ .
February 16th 2009, 02:38 PM
ThePerfectHacker

Quote:

Originally Posted by srw899

A function of the form f(z)=az+b where a and b are complex constants is called a linear transformation. show that every linear transformation can be expressed as the composition of a magnification, a ratation, and a translation. deduce from this that a linear transformation maps lines to lines and circles to circles.

Hint: write a in polar form.

You need to know that $a\not =0$ !

If $a\in \mathbb{R}^{\times}$ and $a>1$ then $z\mapsto az$ is magnification by $a$ . If $0<a<1$ then $z\to az$ is minification by $a$ . If $a<0$ then $a\to az$ is the same effect of $a\to |a|z$ however the object is reflected through the origin. And of course if $a=1$ then nothing happens.

Let $\theta\in \mathbb{R}$ then $z\mapsto e^{i\theta}z$ is a rotation counterclockwise by $\theta$ . If if $\alpha \in \mathbb{C}$ then $z\mapsto z+\alpha$ is a translation in direction of $\alpha$ (if you think of this complex number as a vector).

Therefore, given $f(z) = \alpha z + \beta$ where $f(z) = re^{i\theta} + \beta$ . So first you send $z\mapsto rz$ then $z\mapsto e^{i\theta}z$ and then $z\mapsto z+\beta$ . Thus, all we are doing are rotating, maginifying/minifying, and translating. Thus, lines and circles go to lines and circles, respectively.