# The mapping of complex variable function

So I have the complex variable function $f(Z)=\frac{Z}{Z-1}$ where $|Z|<1$

I have to represent it in $w$-plane. It is how far I got:

$w=\frac{Z}{Z-1} => Z=\frac{w}{w-1}$

and then $|\frac{w}{w-1}|<1 => |w|<|w-1|$

And I dont know how I should go on... As I understand I have to find out what are the possible $w$ values right...?

Am I right taking these steps:

if $w=u+iv$ then

$|w|^2<|w-1|^2$

$u^2+v^2<(u-1)^2+y^2 => u<\frac{1}{2}$

So, the answer is on the left side of graphed line $u=\frac{1}{2}$ ?

Using $|Z|<1$, we know that $|Z-1|\leq |Z|+1$ by the triangle inequality. This will give you an upper bound for $|Z-1|$. Also, note: $$|\frac{Z}{Z-1}|= \frac{|Z|}{|Z-1|}$$. Since you have an upper bound on both of those moduli, then you have an upper bound on their ratio. That is, $$\frac{|Z|}{|Z-1|} < \frac{1}{2}$$ Which is exactly the modulus of $w$. Therefore, $|w|<\frac{1}{2}$, and of course, the lower bound is zero, where the modulus equals zero if and only if $w=0$.
• Hi! thanks for your replay. But I still cant understand what upper bound we get from triangle inequallity? as well as how the modulus ratio is less than $\frac{1}{2}$. Also where from the lower bound is zero? :D – Rsn Apr 5 '15 at 18:19
• Okay, thanks... that was a stupid question about lower bound. So all in all the answer is the area inside of the circle which $r=\frac{1}{2}$ ? – Rsn Apr 5 '15 at 18:39