Let $C_1=\{z:|z|\leq 1, \theta \in [\frac{\pi}{2},\frac{3\pi}{2}] \}$ and $A(z)=z-1$. Define $A(C_1)$. How to show that $A(C_1)\neq\{z:|z-1|\leq 1, \arg(z)=\theta \in [\frac{3\pi}{4},\frac{5\pi}{4}] \}$, but $A(C_1)=\{z:|z-1|\leq 1, \arg(z)=\theta \in [\frac{\pi}{2},\frac{3\pi}{2}] \}$?
My solution for $A(C_1)$:
$A(z)$ is translation, so called elementary function, which translates set $C_1$ one unit to the negative direction of real axis. Thus $A(C_1)=\{z:|z-1|\leq 1, \theta \in [\frac{\pi}{2},\frac{3\pi}{2}] \}$. Comments on my solution? I'm not sure whether my solution is appropriate proof.
The function $A$ is a translation, hence it takes any subset of $\mathbb C$ and it moves it in certain direction; in our case it moves our subset (that is the left half of the closed unit circle) horizontally, shifting it backward of exactly one unit. No more: in particular $A$ is not a rotation, hence the left half of the closed unit circle will remain the left half of a unit circle (but not centered in the origin!), so you won't change the argument of the complex number of the image (under $A$) of your subset: this is why $A(C_1)\neq$ the first set you wrote (because of the argument conditions).
But pay attention! $|z-1|\le1$ denotes the point of the complex plane whose distance from $z_0=1$ is less or equal than $r=1$; so you would have translated your circle forward, not backward! But the translation takes the points backward, so you must consider the point whose distance from $z_0=-1$ is less or equal than $r=1$, i.e. $|z-(-1)|\le1$. Finally you have $$ A(C_1)=\{z\in\mathbb C\;:|z+1|\le1,\;\theta=\arg z\in[\pi/2,3\pi/2]\}\;. $$