Let $A_1=\{z: z(t)=\frac{1}{2}e^{it}- \frac{1}{2}i, t\in [\frac{\pi}{2}, \frac{3\pi}{2}] \}$ and $f(z)=\frac{1}{z}$. Show that $f(A_1)=\{z: z(t)=x+i, x \in \mathbb{R} \}$.
My attempt:
$f(A_1)=f(\frac{1}{2}e^{it}- \frac{1}{2}i)=f(\frac{1}{2}cos(t)- i\frac{1}{2}(sin(t)-1))=\cdots=\frac{cos(t)}{1-sin(t)}+i \neq x+i$. Contradiction. I know that right answer is This $\{z: z(t)=x+i, x \in \mathbb{R} \}$.
Your attempt is almost correct. Notice that $t$ and $x$ are free variable.
It remains to consider $$x(t)=\frac{\cos t}{1-\sin t}$$ has range $\mathbb{R}$ if $t$ is given as stated.