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$$\tan z = \frac{\sin(z)}{\cos(z)}, \cos(z), \sin (z) \ \forall z \in \mathbb{C}$$

Injectivity means: $f(x)=f(y) \Rightarrow x=y$

so i think all intervals of the form $[0,u< 2\pi]$ are good. And I also think this is wrong.

Dose somebody see how to find the correct regions.

Tell me. Please

V

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is $z$ a real number? –  Valerio Capraro Oct 31 '11 at 23:53
    
Anyway, both $sin$ and $cos$ are not injective in that interval. Restrict more. Do you have any idea about their plot? –  Valerio Capraro Oct 31 '11 at 23:55
    
    
great! Now remember that monotonicity implies injectivity and find right subintervals. (Note that $sin$, $cos$ and $tan$ give different results, but I am sure that when you understand the first one, you will be able to do the remaining ones by yourself) –  Valerio Capraro Nov 1 '11 at 0:01
1  
Hint: $\tan(z) = \tan(w)$ if and only if $\tan(z-w) = 0$. And where does $\tan$ take the value 0? –  Robert Israel Nov 1 '11 at 0:45
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