How do i prove sin z is continuous in the complex region?
Thanks Adam
If you mean using the definition of continuity, the same way you would in any region- show that, for any complex number, $\displaystyle z_0$, given any $\displaystyle \epsilon> 0$, there exist a real number $\displaystyle \delta> 0$ so that if $\displaystyle |z-z_0|< \delta$, then $\displaystyle |sin(z)- sin(z_0)|< \epsilon$.
Exactly how you would do that, depends on exactly how you are defining sin(z) for z complex. If you are defining sin(z) in terms of its Taylor series, just subtract the Taylor series for sin(z) and $\displaystyle sin(z_0)$. If you defining sin(z) as $\displaystyle \frac{e^{z)- e^{-z}}{2i}$, are you allowed to use the fact that $\displaystyle e^z$ is continuous?
Or, since sin(z) satisfies the Cauchy-Riemann equations, it follows that it is differentiable and therefore continuous.