I want to "Show that the mapping $f(z) = z + \frac{R^2}{z}$ takes the two concentric circles, $|z|=R$ and $|z| = R'> R$, onto a line segment and an ellipse." (These are depicted in a figure. The line segment goes from $-2R$ to $2R$ on the horizontal axis and the ellipse is "surrounding" it.)
I have access to a solution to the problem, but I don't understand it. There's a factor 2 that comes in, that I don't understand. Here's the given solution:
First let $z=Re^{it}$. This gives $f(Re^{it})=R(e^{it}+e^{-it})=2R\cos t$, which traces the line segment $w=-2R$ to $w=2R$. Here I understand where the factor 2 comes from! But the solution goes on:
Next, consider $f(R'e^{it})=R'e^{it}+\frac{R^2}{R'}e^{-it}=2(R'+\frac{R^2}{R'})\cos t+2i(R'-\frac{R^2}{R'})\sin t$, which is an ellipse surrounding the line segment.
Here, I understand the first step, but where does the factor of 2 come in, in the second step? I get the same expression but without the factor 2.
Probably I'm making a stupid mistake somewhere that I can't see. Thankful for any help with this!
The factor of two should not be there.
$f(R'e^{it})=R'e^{it}+\frac{R^2}{R'}e^{-it}=(R'+\frac{R^2}{R'})\cos t+i(R'-\frac{R^2}{R'})\sin t$
The ellipse contains the line segment - consider $R' \to R$
$$R' + \frac{R^2}{R'} \to R + \frac{R^2}{R} = R + R = 2R$$