Image of a circle under a complex valued function Find the image of $|z-1| = 1$ under the transformation $w = (1+i)z - 2$.  
This is confusing as I just started out in this topic.
I drew the circle $(x-1)^2 + y^2 = 1$. I see that the equation first wants me to rotate and extend the length of each point (w.r.t its connection to the origin) of the circle by $\frac{\pi}{4}$ radians and by a multiple of $\sqrt{2}$ respectively. Then I shift the circle $2$ units to the left.
However, I'm having trouble finding the equation for this.  
I would prefer a purely algebraic way to manipulate it into the equation.
 A: $w(z)=(1+i)z-2=(1+i)z-(1+i)-2+(1+i)=(1+i)[z-1]+(1-i)$, hence
$|w(z)-(1-i)|=|1+i||z-1|= \sqrt{2}$ for $|z-1|=1$.
A: Steps:


*

*You start with a circle centered at $1$ with radius $1$.

*Then you apply to it a clockwise rotation of $\frac\pi4$ radians; you get a circle centeres at $\frac{1+i}{\sqrt2}$ with radius $1$;

*Then you apply a homothety centered at $0$ with reason $\sqrt 2$; you get a circle centeres at $1+i$ with radius $\sqrt2$.

*Finally, you shift everything by $2$ units to the left; you get a circle centered at $-1+i$ with radius $\sqrt2$.


Yes, you had already written most of it. What's the problem with this approach? Why do you you feel that you need an algebraic manipulation of the expression?
A: Note that $T(z)=(1+i)z−2=e^{i\pi/4}(\sqrt{2}z)-2$ so it is a  dilation of $\sqrt{2}$ composed with a clockwise rotation of $45^{\circ}$ and then a translation to the left by $2$.
Therefore 
$$|z−1|=1\rightarrow |z−\sqrt{2}|=\sqrt{2}\rightarrow |z−(1+i)|=\sqrt{2}\rightarrow |z−(-1+i)|=\sqrt{2}.$$
