# Thread: 1 sum. elementary transformation from z-plane to w-plane (complex number)

1. ## 1 sum. elementary transformation from z-plane to w-plane (complex number)

For the transformation $\displaystyle w=z^2$ show that as z moves once round the circle centre O and radius 2, w moves twice round the circle center O and radius 4.

I have two problems for which I am stuck on the sum:

1. How do moving "once" and "twice" is represented on the equation? That is, if w lies on circle center O and radius 4, $\displaystyle |w|=4$, but how do show that it moves twice?

2. If i take $\displaystyle |z|=2, w=z^2, z=\sqrt w, |z|=|\sqrt w|=2$: how do I do this part: $\displaystyle 2=|\sqrt w|$, when w is a complex number?
And how do i carry out from there :help:

For the transformation $\displaystyle w=z^2$ show that as z moves once round the circle centre O and radius 2, w moves twice round the circle center O and radius 4.

I have two problems for which I am stuck on the sum:

1. How do moving "once" and "twice" is represented on the equation? That is, if w lies on circle center O and radius 4, $\displaystyle |w|=4$, but how do show that it moves twice?

2. If i take $\displaystyle |z|=2, w=z^2, z=\sqrt w, |z|=|\sqrt w|=2$: how do I do this part: $\displaystyle 2=|\sqrt w|$, when w is a complex number?
And how do i carry out from there :help:
The path that you take on the circle at origin of radius two can be expressed as $\displaystyle g(\theta) = 2e^{i\theta}$ for $\displaystyle 0\leq \theta \leq 2\pi$.
Under the transformation $\displaystyle z\mapsto z^2$ the path becomes mapped to $\displaystyle \left( 2e^{i\theta} \right)^2 = 4e^{2i\theta}$ for $\displaystyle 0\leq \theta \leq 2\pi$. Because of the presence of $\displaystyle 2i\theta$ (rather than $\displaystyle i\theta$) it means the points moves twice around a circle of radius 4.

3. Originally Posted by ThePerfectHacker
The path that you take on the circle at origin of radius two can be expressed as $\displaystyle g(\theta) = 2e^{i\theta}$ for $\displaystyle 0\leq \theta \leq 2\pi$.
Under the transformation $\displaystyle z\mapsto z^2$ the path becomes mapped to $\displaystyle \left( 2e^{i\theta} \right)^2 = 4e^{2i\theta}$ for $\displaystyle 0\leq \theta \leq 2\pi$. Because of the presence of $\displaystyle 2i\theta$ (rather than $\displaystyle i\theta$) it means the points moves twice around a circle of radius 4.
Gotcha, thanks