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$\begingroup$

My attempt

$$f(z)= \frac{\cos z}{(z-1)}$$

$$f^{\prime}(z)= -\frac{\sin z}{(z-1)}-\frac{\cos z}{(z-1)^2}$$

$$\int \frac{cos(z)}{z^2(z-1)}dz=2\pi if^{\prime}(0)=2\pi i\left( -\frac{\sin 0}{(0-1)}-\frac{\cos 0}{(0-1)^2}\right)= 2\pi i\left(-\frac{\sin 0}{(0-1)}-\frac{\cos 0}{(0-1)^2}\right)=-2\pi i$$

Is this correct?

  • 0
    By $|z|<\frac{1}{3}$ do you mean a contour around there? – 2017-01-16
  • 1
    It seems to me fine. – 2017-01-16
  • 0
    @Mark, Yes it is. – 2017-01-16
  • 0
    looks OK $\quad{}$ – 2017-01-16

0 Answers 0