large possible regions where $\tan z , \sin z , \cos z$ are injective

Asked Oct 31 '11 at 23:46

Active Nov 01 '11 at 00:17

Viewed 232 times

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

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

edited Nov 01 '11 at 00:17

Brian M. Scott

asked Oct 31 '11 at 23:46

VVV

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
http://www.wolframalpha.com/input/?i=cos%28z%29+ ; http://www.wolframalpha.com/input/?i=sin%28z%29+ – VVV Oct 31 '11 at 23:57
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 01 '11 at 00:01
OK, i restrict it like this: $[Re(z) > 0,Im(z)>0]$ and the other half is $[Re(z) < 0 , Im(z)< 0 ]$ – VVV Nov 01 '11 at 00:21
is this correct for cosz? – VVV Nov 01 '11 at 00:35
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 01 '11 at 00:45

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