$\oint_{\left | z \right |=0.5} \frac{dz}{(z-1)(\sin z)} $

Define: $f(z) = \frac{z}{(\sin z)(z-1)}$

Define: $g(z) = \frac{f(z)}{z}$

Now integrate using Cauchy Integration Formula

$\oint_{\left | z \right |=0.5} g(z) dz=2i\pi f(0)=-2i\pi $

where the minus sign comes from the following limit:

$\lim_{z \to 0}f(z) = \frac{0}{0} = \frac{1}{(\cos z)(z-1) + \sin z}=-1$

I see that zero is a removable discontinuity point for f(z). Under these conditions may i also apply the Cauchy Integral Formula? Is the result correct?

  • 1
    $\begingroup$ It seems you're looking at this question. $\endgroup$
    – Pedro
    Commented Dec 21, 2014 at 15:40

1 Answer 1


Looks like you are using L'Hospital's Rule in the last step, which doesn't apply in the complex setting. However, you can show that

$$\lim_{z\to 0}\frac{\sin z}{z}=1$$

by using the Taylor expansion of $\sin$.

Other than that, your answer looks fine.


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