Apply the Cauchy-Goursat Theorem to show that $$\oint_C f(z) \ dz = 0$$ when the contour $C$ is the unit circle $|z|=1$, in either direction, and when $$ f(z)=tan(z)$$

Reminder: Cauchy-Goursat Theorem says, If a function f is analytic at all points interior to and on a simple closed contour $C$, then $\oint_C f(z)\ dz =0$

I'm aware that this function is not entire and that the problem occurs when $cos(z)=0$, so why does Cauchy-Goursat still apply here?

  • $\begingroup$ $\tan(z)$ has first-order poles at $z=(2n+1)\pi/2$ for every $n$. $\endgroup$ – Mark Viola Nov 3 '16 at 22:02
  • 1
    $\begingroup$ Briefly: $\frac{\pi}{2} > 1$. The poles of $\tan$ are outside the region of interest, $\tan$ is analytic on a neighbourhood of the closed unit disk. $\endgroup$ – Daniel Fischer Nov 3 '16 at 22:19
  • $\begingroup$ We never get $\cos(z)=0$ inside $C.$ $\endgroup$ – Simply Beautiful Art Nov 3 '16 at 22:25

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