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 $f$.
Best Answer
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 $|w-\frac14|^2 = \frac1{16}$.