Find a mobius transfromation that maps $|z|=2$ into $|z+1|=1$ Find a mobius transfromation that maps $|z|=2$  into $|z+1|=1$. mapping $-2$ and $0$ to $0$, $i$ respectively. 
I started by substituting in the general form  $T(z)= \frac{az-b}{cz-d}$. 
I get $b = 2a $and $ d =-ib$. Now, since $T(0) \neq \infty $ we set $d=1$ which gives $b=i$ and
$a=\frac{1}{2}i$. 
Next thing I know is that 
$| \frac{\frac{1}{2}iz+i}{-2c+1}-1|=1$ when $|z|=2$
I want to use it to find $c$ but I don't know how to proceed and I'm not sure if my previous steps are valid.
 A: First I’ll tell you what the solution to the problem is, then I’ll check that it’s correct, then I’ll show you the (inefficient) method I used.
The Solution:
$$
T(z)=i\frac{z+2}{(1-i)z+2}
$$
Check the Result:
We need. for instance, the three points $\pm2$, $2i$ to be sent to points on the circle centered at $-1$ of radius $1$, and in particular for $-2$ to be sent to the origin. This last demand is obviously satisfied, so let’s look at $z=2$: goes to $4i/(4+2i)=\frac25+\frac45i$. Add $1$ and you get a point obviously on the unit circle. Finally, try $T(2i)=(2+2i)/(4+2i)=-\frac15+\frac35i$, again clearly on the unit circle when you add $1$. Only remains to check that $0$ goes to $i$, and that’s obvious.
How I got it:
I did not use your method. I’ll use the familiar device of representing a Möbius transformation $\frac{az+b}{cz+d}$ by a two-by-two matrix $(a,b;c,d)$, so that composition of Möbii corresponds to product of matrices, earliest operation at the right.
My strategy was to shrink the first circle to unit radius and move the interesting point $-2$ to $+1$, do my transformations on that unit circle while leaving $1$ fixed, and then shift over by $1$ unit to the left. Unfortunately, I am not so skilled that I can tell you what a typical circle-preserving, $1$-fixing Möbius looks like. I can only transform the circle to the Upper Half Plane (UHP) while sending $1$ to $\infty$. There, in UHP, I know exactly what the transformations are that leave $\infty$ fixed: they’re just $z\mapsto\lambda z+\mu$ for positive $\lambda$ and real $\mu$. That matrix is $(\lambda,\mu;0,1)$. Before we start working, I have to point out to you that ultimately we’re sending the origin inside  our radius-$2$ circle, to $i$, outside our radius-$1$ circle. So there’s going to have to be something like $z\mapsto1/z$ acting on the unit circle to do that. Fortunately, $1$ is fixed there.
Now let’s get to work. The last operation we do is left-shift by $1$, and to get our original origin to wind up at $i$, all the steps before the last must get the o.o. to $1+i$. If, just before that we’ve turned the disk inside out, we must start with a point that inverts to $1+i$, namely $\frac12-\frac12i$. We have almost all the information we need, except the description of how to turn the unit disk into UHP so that $1\mapsto\infty$.
While we’re doing that, we might as well send $-1\mapsto0$ and $0\mapsto i$. That’ll put the real axis inside the circle to the positive imaginary axis.
$$
\text{send disk to UHP:}M=\begin{pmatrix}1&1\\i&-i \end{pmatrix}
\qquad
\text{its inverse:}M^{-1}=\begin{pmatrix}i&1\\i&-1\end{pmatrix}
$$
What point in the plane corresponds to $\frac12-\frac12i$ in the disk? Apply $M$ to that number, and get $2+i$. Amazing! All we need to do in the plane is shift everything rightwards by $2$ to send $i$ (the image of the original origin) where we want it. So our steps are (shrink original circle by $2$ and turn), (then push it all into UHP), (move right $2$), (push back into unit disk), (turn inside-out), (finally, shift left $1$).
In order, then remembering to read right-to-left:
$$
\begin{pmatrix}1&-1\\0&1\end{pmatrix}
\begin{pmatrix}0&1\\1&0\end{pmatrix}
\begin{pmatrix}i&1\\i&-1\end{pmatrix}
\begin{pmatrix}1&2\\0&1\end{pmatrix}
\begin{pmatrix}1&1\\i&-i\end{pmatrix}
\begin{pmatrix}-1&0\\0&2\end{pmatrix}
$$
Multiply them all out, and remove any common factors from all four entries, and get \begin{pmatrix}i&2i\\1-i&2\end{pmatrix}
which is what I proclaimed at the top.
A: Professor Lubin’s answer demonstrates a lot of skills in doing the mapping. Here are two other approaches. To begin, let $T(z)$ be the required mapping. By the given conditions, it is clear that $T(z)$ is of the form $$T(z)=\frac{i(z+2)}{cz+2}$$ for some $c\in{\mathbb C}.$
First approach: To solve for $c$, assume that $c=a+bi$ with $a,b\in{\mathbb R}.$ Then by straightforward computation, one has $$|T(2)+1|=1\Rightarrow b=-1,{\rm ~so~}c=a-i.$$ Furthermore, $$|T(2i)+1|=1\Rightarrow a=1.$$ Thus $c=1-i$. It remains to check that this works, namely one checks that $$|T(z)+1|=1~{\rm whenever~}|z|=2$$ $$\Leftrightarrow \left|\frac{z+2(1+i)}{(1-i)z+2}\right|=1,$$ which is checked directly by expanding the square of the norm in the numerator and the denominator and using $|z|^2=z\overline{z}=4.$ Note that the proof shows that $T$ is uniquely determined.
Second approach: Note that $T(\infty)\neq \infty$, since $c=0$ implies that $|T(2)+1|\neq 1.$ It follows that $T(\infty)$ is a finite value, so $T(-2),T(0),$ and $T(\infty)$ defines a circle $C$ orthogonal to the circle $|z+1|=1,$ by the conformal property of Möbius transformation, as the $x$-axis is orthogonal to the circle $|z|=2.$ Since $C$ passes through $0,i$ and is orthogonal to $|z+1|=1,$ $C$ must have the equation of the form $$x^2+\left(y-\frac 12\right)^2=\frac 1 4.$$ Now circle $C$ intersects $|z+1|=1$ (i.e. $(x+1)^2+y^2=1$) at $(0,0)$ and $\left(-2/5,4/5\right).$ Necessarily $(-2/5,4/5)$ must be the image of $(2,0)$ under $T,$ i.e. $$T(2)=-\frac 2 5+\frac 4 5 i,$$ which yields $c=1-i.$ Now one checks that $T(-2),T(2)$ and $T(2i)$ pass through $|z+1|=1$, hence $T(z)$ with $c=1-i$ is the required mapping. QED
