Let [imath]f:\mathbb{C} \setminus \{-1\} \to \mathbb{C} \setminus \{1\}[/imath], let [imath]A=\{z\in\mathbb{C} \ | \ \Re(z) >0\}[/imath] and let [imath]D=\{z\in\mathbb{C} \ \text{such that} \ |z| <1\}[/imath]. Prove that [imath]f[/imath] maps [imath]A[/imath] in [imath]D[/imath].
My work: let [imath]z \in A[/imath] be arbitrary. Letting [imath]z=x+iy[/imath], it is:
[math]|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}}[/math]Since [imath]z \in A[/imath], [imath]x>0[/imath] and so [imath]-2x<2x[/imath]. Using the facts that the square root is increasing and that the denominator of the latter fraction is positive, we have:
[math]\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[/math]So, [imath]|f(z)|<1[/imath]; hence, [imath]f(A) \subseteq D[/imath]. Is this correct?
My work: let [imath]z \in A[/imath] be arbitrary. Letting [imath]z=x+iy[/imath], it is:
[math]|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}}[/math]Since [imath]z \in A[/imath], [imath]x>0[/imath] and so [imath]-2x<2x[/imath]. Using the facts that the square root is increasing and that the denominator of the latter fraction is positive, we have:
[math]\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[/math]So, [imath]|f(z)|<1[/imath]; hence, [imath]f(A) \subseteq D[/imath]. Is this correct?
yes, the function law is [imath]f(z)=\frac{z-1}{z+1}[/imath].