Determine the image of the infinite strip $1 ≤ \Re(z) ≤ 2$ under the transformation $f(z) = 1/z$

Question

Determine the image of the infinite strip $1 ≤ \Re(z) ≤ 2$ under the transformation $f(z) = \frac 1z$ and also determine the image of the infinite strip $0 ≤ \Re(z) ≤ 2$ under $f$.

My attempt for $\Re(z)=2$ ; $f(z) =u(x,y)+iv(x,y)$

Given $z=2+iy$, then $f(z)=\frac 1 {2+iy}$, thus $f(z)=\frac{2-iy}{4+y^2}$

$f(z) =u(x,y)+iv(x,y)$ ; $u(x,y)=\frac 2 {4+y^2}$ and $v(x,y)=\frac {-y} {4+y^2}$

This eventually gives us $(x− \frac 1 4)^2+y^2=(\frac 1 4)^2$ ; circle of radius $\frac 1 4$ and is centered at $(\frac 1 4 , 0)$

I'm not sure where to continue with this question. Should I evaluate the transformation at $\Re(z)=0$ ? - This would be $u(x,y)$ = 0 and $v(x,y)=\frac{-1}{y}$ correct?

I could really do with some help with this, and I assume I would use the same method for the second part to determine the image of the infinite strip $0 ≤ \Re(z) ≤ 2$ under $f$.

Answer 1

Let $w = \frac1z$ and $Re(z)=a$. Then, we have $z+\bar z = 2a$ and $\frac1w+\frac1{\bar w}=2a$, which leads to

$$2a|w|^2 - (w+\bar w) = 0\implies |w-\frac1{2a}|^2=\frac1{4a^2}$$

So, for any given $a$, $w(z)$ represents a circle of the center $\frac1{2a}$ and the radius $\frac1{2a}$. Therefore, $1\le a \le 2$ generates the image enclosed by two circles,

$$|w-\frac14|^2 = \frac1{16},\>\>\>\>\>|w-\frac12|^2 = \frac14$$

as shown in the graph below,

Similar, for $0<a\le 2$, the outer circle becomes infinitely large. Then, the image is the left half plane, minus the circlar area $|w-\frac14|^2 = \frac1{16}$.

Answer 2

You can read very detailed explanation about generalized circles and the $f(z) = \frac{1}{z}$.

The basic take-home are:

• lines through origin are mapped to lines through origin (specifically, $\Re(z) = 0$ will map to itself, and $\mathbb{R}$ will also map to itself)
• lines not through origin will be mapped to circles through origin.

$\Re(z) = 1$ will be mapped to a circle through origin and $1$, i.e. (since it is symmetric with regards to $Im(z) = 0$) a circle centered at $(\frac{1}{2}, 0)$ with radius $\frac{1}{2}$.

By the same reasoning $\Re(z) = 2$ will be mapped to a circle through $(0,0)$ and $(\frac{1}{2}, 0)$ symmetric with respect to real axis (i.e. a circle at $(\frac{1}{4}, 0)$ with radius $\frac{1}{4}$ (as you stated).

Now, the strips will be between appropriate circles. In the $0 \leq \Re(z) \leq 2$ - half a plane without the circle!.