Image of a complex valued function: Let f : C \ {-1} -> C \ {1}, A = {z in C | R(z) > 0}, and...

[Ozma]

Let $f:\mathbb{C} \setminus \{-1\} \to \mathbb{C} \setminus \{1\}$, let $A=\{z\in\mathbb{C} \ | \ \Re(z) >0\}$ and let $D=\{z\in\mathbb{C} \ \text{such that} \ |z| <1\}$. Prove that $f$ maps $A$ in $D$.

My work: let $z \in A$ be arbitrary. Letting $z=x+iy$, it is:
$$|f(z)|=\left|\frac{z-1}{z+1}\right|=\frac{|z-1|}{|z+1|}=\frac{\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}}=\frac{\sqrt{x^2-2x+1+y^2}}{\sqrt{x^2+2x+1+y^2}}$$Since $z \in A$, $x>0$ and so $-2x<2x$. Using the facts that the square root is increasing and that the denominator of the latter fraction is positive, we have:
$$\frac{\sqrt{x^2-2x+1+y^2}}{\sqrt{x^2+2x+1+y^2}}<\frac{\sqrt{x^2+2x+1+y^2}}{\sqrt{x^2+2x+1+y^2}}=1$$So, $|f(z)|<1$; hence, $f(A) \subseteq D$. Is this correct?

 

[blamocur]

Looks good to me. It would be easier to understand if the problem statement included $f(z) = \frac{z-1}{z+1}$.

 

[Ozma]

@blamocur: Thank you! I forgot the most imporant information yes, the function law is $f(z)=\frac{z-1}{z+1}$.