How to compute a linear transformation which carries the circle $|z|=2$ into $|z+1|=1$? 
Find the linear transformation which carries the circle $|z|=2$ into $|z+1|=1$, the point $-2$ into the origin, and the origin into $i$. 

In order to find the linear transformation will I use the following formula?
$$(z_1−z_3)(z_2−z)(z_2−z_3)(z_1−z)=(w_1−w_3)(w_2−w)(w_2−w_3)(w_1−w)$$ 
Or am I wrong?
Similar example but not the same:

Find a linear transformation mapping the circle $|z|=1$ onto the circle $|w-5|=3$ and taking the point $z=i$ to $w=2$. 

Given, $|w-5|=3$ so $|(w-5)/3)|=1$. So now, we can use the transformation $z=(w-5)/3$ but $z=i$ is not mapped to $w=2$. So we use the transformation $iz=(w-5)/3$. So, $z=i \implies w=3i*i+5=2$.
How to start ?
 A: The linear transformation you are looking for is of the form:
$$
\varphi(z)=\frac{\omega z}{2}-1,
$$
where $\lvert\omega\rvert=1$.
Verification. If $\lvert z\rvert=2$, then $z=2\mathrm{e}^{it}$, and $\varphi(z)=\omega\mathrm{e}^{it}-1$, which is the parametrical description of the circle $\lvert z+1\rvert=1$.
Determination of $\omega$. It is going to be determined by the restriction
$$
\varphi(-2)=0,
$$
equivalently
$$
0=\frac{\omega(-2) }{2}-1=-(1+\omega),
$$
and hence
$$
\omega=-1.
$$
Thus the linear transformation is 
$$
\varphi(z)=-\frac{z}{2}-1,
$$
Clearly $\varphi(0)=-1$. 
Apparently, such linear transformation does not exist.
A: Any complex-linear transformation has the form $z \mapsto az + b$. You only need to find $a$ and $b$.
There are at least two ways to proceed: geometrically, you can work out the effect of the transformation in terms of how much it scales, rotates, and translates the plane, and then use the fact that multiplication by $re^{i\theta}$ scales by $r$ and rotates by $\theta$, and translation is just addition of a constant.
Alternatively, you can proceed algebraically: eliminate $b$ by considering the effect of the transformation on the difference between two points. If you know $w$ transforms to $x$ and $z$ transforms to $y$, then $x - y = (aw + b) - (az + b) = a(w - z)$. This gets you $a$, and then you can solve for $b$.
As Yiorgos notes, there doesn't actually seem to be a solution in this case. One quick way of seeing that is to observe that the origin travels from inside the circle to outside of it. Linear transformations just can't do that.
A: Not sure if this is correct so please let me know
Let $C_1 = |z| = 2$, and $C_2 = |z + 1| = 1$. Since the origin is in $C_1$ but not in $C_2$, we need to make the circle go inside out. Thus we get $T = \frac{az + b}{cz + d}$.
We know $T = z*(\frac{z + b}{z + d})$ $z*$ be the reflection point of the origin. We let $R$ be the radius of $C_2$, and $\alpha$ be the center of $C_2$.
Now $\infty$ is the inflection point of $0$. Therefore $Z^* = \frac{R}{\bar{z} - \bar{\alpha}} + \alpha = \frac{1}{-i + 1} -1 = \frac{1 + i - 1}{-i + 1} = \frac{i}{1 - i} = \frac{-1}{1 - i} \times \frac{1 + i}{1 + i} = \frac{-1 +i}{1 + 1} = \frac{-1 + i}{2}$.
So $T = \frac{-1 + i}{2} \times \frac{z + b}{z + d}$. So since $T(-2) = 0$ we get $b = 2$. It remains to solve for $d$
. We get $T(0) = i = \frac{-2 + 2i }{2d}$ therefore $d = \frac{-1 + i}{i} = i + 1$. 
So $T = \frac{(-1 + i)z + 2}{2z + (i + 1)}$
