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I know that $f$ and $g$ have a pole or order $k$ in $z=0$. $f-g$ is holomorph in $\infty$.

I need to prove that:

$$\oint_{|z|=R} (f-g)' dz = 0$$

Any help?

Note: $f$ and $g$ only have a singularity in $z=0$

Git Gud
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Unnamed
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2 Answers2

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As Git Gud alluded to in their comment, simply parameterise the curve and calculate the integral directly. The usual parameterisation is $\gamma:[0,2\pi]\rightarrow\mathbb C:\ t\mapsto Re^{it}$. Letting $h=f-g$, we have $$\oint_Ch'(z)\ \mathrm dz=\int_0^{2\pi}h'(\gamma(t))\gamma'(t)\ \mathrm dt=\int_0^{2\pi}(h\circ\gamma)'(t) \mathrm dt\\=(h\circ\gamma)(2\pi)-(h\circ\gamma)(0)=h(R)-h(R)=0.$$

Jason
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If $h$ is holomorphic, then $h'$ exists and is holomorphic. If $h$ is holomorphic, then $\oint_C h dz = 0$ for all circles $C$. Take $h= f-g$, $(f-g)'$ respectively.

nullUser
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  • No, the exercise is not so stupid haha. $f-g$ is holomorphic in $z=\infty$ – Unnamed Feb 20 '15 at 02:04
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    What does "holomorphic in $z=\infty$" mean? I know of holomorphic on an open set. – nullUser Feb 20 '15 at 02:05
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    This isn't quite true. If $h$ is holomorphic on a region $D$ and $C$ is a closed simple curve entirely within $D$, then $\oint_Ch(z)\ \mathrm dz=0$. But $h=(f-g)'$ is not holomorhpic at $z=0$. – Jason Feb 20 '15 at 02:35
  • @Jason why is that? the OP did not say that $f-g$ is not holomorphic at $z=0$. He said that it is holomorphic at $\infty$ which I presume, maybe mistakenly, that it's entire. For example, take $f=g=\frac{1}{z}$ – benji Feb 20 '15 at 22:07
  • Well fine, it is not necessarily holomorphic at $z=0$. Your example doesn't change the fact that we do not know for sure that $(f-g)'$ is holomorphic at $z=0$, so we cannot use Cauchy's theorem. – Jason Feb 21 '15 at 02:12