Studying the complex-valued function $f(z) = \frac{1}{z}$ Let $f(z) = \frac{1}{z}$. I am trying to study this function: First thing to notice is that we can write (after some algebraic manipulation and putting $z = x + i y $) that 
$$ f(x,y) = \frac{x}{x^2 + y^2} - \frac{y}{x^2+y^2} i $$
Let $D = \{ z : |z| < 1 \} $ and $E = \{ z : |z| > 1 \} $
I know that $f(D)$ is also the unit circle since if we take some $z \in D$, then if $z= (x,y)$ we have $x^2 + y^2 < 1 $ and
$$ |f(z)| = |f(x,y)| = \bigg| \frac{x^2 + y^2}{(x^2+y^2)^2} \bigg| < 1 $$
question: IS the same true for the set $E$? In other words, the image of $E$ under $f$ is $E$ again? thanks
 A: It is not true that $f(D)$ is in $D$. (Also, note that you have $1/0$!) Yes, $x^2 + y^2 < 1$, but you divide by it. Likewise it is not true that $f(E)$ is in $E$.
In fact $f(D\setminus \{0\})= E$ and $f(E)= D \setminus \{0\}$, the sets are interchanged, and $f(U) = U$ where $U$ denotes the complex numbers of absolute value $1$. 
But best forget about real and imaginary part in this case. Recall that $|1/z| = 1/|z|$.
A: $f$ is an example of a Mobius transformation. It is also closely related to the circle inversion map $g(z) = \frac{1}{\overline{z}}$. Geometrically (i.e. as long as you don't care about analyticity) it is slightly more convenient to talk about $g$ than $f$, but everything I say here applies to both.
$g$ has the property that it sends most lines to circles and most circles to lines. The exceptions are lines through the origin, which are fixed under $g$, and circles centered at the origin, which go to circles centered at the origin of reciprocal radius. (So the unit circle is also fixed.)
This ought to give you an idea of how $f$ acts on $D$ and $E$; think of them as unions of circles centered at the origin.
