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I have show that the function $\sin(z)$ satisfies the Cauchy-Riemann equations but don't know where to go from here.

If it saves some working for you they are

$$du/dx=-\sin(x)\cosh(y)$$ $$dv/dy=-\sin(x)\cosh(y)$$

$$du/dy=\cos(x)\sinh(y)$$ $$dv/dx=-\cos(x)\sinh(y)$$

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    Do you know what "analytic" means?2012-11-26
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    Alternatively, $\sin(z) = -\frac{i}{2}(e^{iz} - e^{-iz})$ is obtained from analytic functions by addition, composition, and multiplication, hence is analytic.2012-11-26
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    Since the partial derivatives are continuous, the function is also real-differentiable. Thus, it's analytic.2012-11-26
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    I don't understand: if you wanted to prove $\,\sin z\,$ is analytic then you already did the hard work with those partial derivatives! What else do you want? Do you know what "analytic" means?2012-11-26
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    I haven't been given a definition anywhere but I do have the theorem that "if the partial derivates of f exist and are continous on $omega$ and the Cauchy Riemann differential equations are satisfied there, then f is analytic on $omega$"2012-11-26
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    Do I just need to state that sin(z) is a continous function?2012-11-26
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    @ChrisEagle sorry forgot to tag you in my comment2012-11-26

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