The map $w=f(z)=\frac{i(1-z)}{1+z}$. I am now studying complex analysis now. Now I want to find the image of right half plane $Re(z)>0$ under the linear transformation $w=f(z)=\frac{i(1-z)}{1+z}$.
 A: One thing that's useful to know about functions of the form $f(z)=\dfrac{az+b}{cz+d}$ is that they preserve circles on the Riemann sphere.  The Riemann sphere is $\mathbb C\cup\{\infty\}$, the plane with one point at infinity ajoined, where you can approach that point by going in any direction at all.  A circle on the Riemann sphere is either a circle in the usual sense or a straight line.  When it's a straight line, then $\infty$ is one of the points on the circle.
The boundary of the right half-plane is a straight line, therefore a circle on the Riemann sphere; so its image under this mapping is either another straight line or a circle.
So notice that
\begin{align}
f(0) & = i \\
f(i) & = 1 \\
f(-i) & = -1 \\
f(\infty) & = -i 
\end{align}
These arguments to $f$ are on the "circle" that is the imaginary axis, and these four values of $f$ are on a different common circle.  That circle is the boundary of the image of the right half-plane.  Next, you need to figure out which side of the boundary is the image of the interior of the right half-plane.  Is $0$ in that image, or is $\infty$ in that image?
A: Here is a partial map of a region in the right hand plane.

The three curved arcs at the bottom are the maps of the upper, lower and right hand boundaries. I believe the complete map of the right hand plane would be something like what is called a Smith Chart in electrical engineering. Although there they usually map the upper half plane.
Since the Riemann sphere mapping is mentioned in Michael Hardy's answer here is that map. If you could rotate the sphere you would obtain a better idea of the complete mapping.

A: Let's take a look at the boundary of the right half-plane $D=\{z\mid \mathcal{Re}\ z>0\}$, which is the the $OY$-axis that can be parametrized by $t$ as $$z=it, t\in\mathbb{R}$$ and let us investigate how the boundary is transformed under $f$:
$$f(it)=i\frac{1-it}{1+it}=\ldots=\underbrace{\frac{2t}{1+t^2}}_{u(t)}+i\underbrace{\frac{1-t^2}{1+t^2}}_{v(t)}=u(t)+iv(t)$$
Notice that $$|w|^2=|f(it)|^2=u^2(t)+v^2(t)=\frac{4t^2}{(1+t^2)^2}+\frac{(1-t^2)^2}{(1+t^2)^2}=\ldots=1$$ Therefore the image of the boundary is a part of the unite circle $|w|=1$. However, it is easy to see that $1\le u(t)\le1$ and $1\le v(t)\le1$ if we accept that $t\in\mathbb{R}\cup\{\infty\}$. Thus, under $z\stackrel{f}\to w$ the boundary transforms into the unit circle $|w|=1$.
Since the boundary is mapped into the boundary and the domain $D$ will either be inside or outside of the unit circle. In order to determine this, let's pick a simple test point in $D$. For example, $z=1+i0\in D$.
$$f(1+i0)=0. $$
Since $f(1+i0)$ is inside the unit circle, the entire half-plane must be mapped into the interior of the same circle, and we arrive to the conclusion: $$f(D)=\{w\in\mathbb{C}:| w|<1\}.$$

