# Complex Mappings

• Nov 18th 2010, 04:36 PM
dwsmith
Complex Mappings
According to the Remarks in Section 2.5, since $\displaystyle f(z)=\frac{1}{z}$ is its own inverse function, the mapping $\displaystyle w=\frac{1}{z}$ on the extended complex plane maps the circle $\displaystyle |z-\frac{1}{2}|=\frac{1}{2}$ to the line Re(w) = 1. Verify this.

I am not sure how to do this without working backwards which isn't an easy method. Does anyone know how this can be done in a more efficient manner?

Dustin.
• Nov 18th 2010, 09:52 PM
CaptainBlack
Quote:

Originally Posted by dwsmith
According to the Remarks in Section 2.5, since $\displaystyle f(z)=\frac{1}{z}$ is its own inverse function, the mapping $\displaystyle w=\frac{1}{z}$ on the extended complex plane maps the circle $\displaystyle |z-\frac{1}{2}|=\frac{1}{2}$ to the line Re(w) = 1. Verify this.

I am not sure how to do this without working backwards which isn't an easy method. Does anyone know how this can be done in a more efficient manner?

On the circle you may put $\displaystyle z=\frac{1}{2}(1+e^{i\theta}), \theta \in [0,2\pi)$
Now compute $\displaystyle w=\frac{1}{z}$ and show that its real part is equal to $\displaystyle $$1 and its imaginary part takes every real value for some \displaystyle$$ \theta \in [0,2\pi)$