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I'm trying to verify Cauchy's integral theorem "by hand" for $$ I = \oint_C\frac{z}{2z-5}dz\,, $$ where $C$ is the circle $\vert z-3\vert=2$. From Cauchy's integral theorem, the answer is $5\pi i /2$ although I'm not quite sure how to finish the last step of my verification.

My attempt is as follows. \begin{equation*} \begin{aligned} I &= \oint_C \frac{z-3}{2(z-3)+1}dz+\oint_C\frac{3}{2(z-3)+1}dz \\ &= \int_0^{2\pi i} \frac{e^{2x}}{e^x+1/4}dx + \frac{3}{2}\int_0^{2\pi i} \frac{e^{x}}{e^{x}+1/4}dx \\ &= \int_0^{2\pi i} \frac{e^x(e^x+1/4-1/4)}{e^x+1/4}dx + \frac{3}{2}\left[\log{\left(e^{ix}+1/4\right)}\right]_0^{2\pi i} \\ &= \left[e^{x} + \frac{5}{4}\log{\left(e^{x}+1/4\right)}\right]_0^{2\pi i} \\ &= \frac{5}{4}\log{\left(e^x+1/4\right)}\bigg\vert_0^{2\pi i} \end{aligned} \end{equation*} How do I show that the evaluation of the log term is $2\pi i$? The factor of $1/4$ in the $\log$ is throwing me off. I suspect that I am to use $$ \log z = \log r + i(\theta+2\pi n) $$ here, although if I am, I feel silly because I'm not seeing a simple way to employ it.

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  • $\begingroup$ $\ln(e^x + 1/4)$ is discontinuous on the segment $[0, 2 \pi i]$. You can evaluate your integral without a change of variables as $$I = \frac 5 4 \ln(2 z - 5) \bigg\rvert_{z = 1 - i0}^{1 + i0},$$ where $\ln$ is the principal branch. $\endgroup$
    – Maxim
    Oct 29 '20 at 16:59
  • $\begingroup$ I think this is more along the lines of the answer I'm looking for, although I have a clarification question. I thought you were putting the bounds with the $i0$ terms to emphasize that the evaluation is to be done over different adjacent branches of $\ln$, giving this extra term of $2\pi i$. However, you seem to be emphasizing at the end that I am supposed to evaluate both over the principal branch? If so, I'm confused where I'm going to get the $2\pi i$ from... $\endgroup$
    – Hale Bays
    Oct 30 '20 at 1:21
  • $\begingroup$ $f(z + i0)$ denotes the limit from above (in the complex plane) at $z$. $\endgroup$
    – Maxim
    Oct 30 '20 at 2:19
  • $\begingroup$ Okay. Also, did you instead mean $z=5\pm i0$, given that $C$ is the circle $\vert z-3\vert = 2$? $\endgroup$
    – Hale Bays
    Oct 30 '20 at 5:00
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Write $I = \oint_C\frac{z}{2z-5}dz=\frac12\oint_C \frac{z-\frac52+\frac52}{z-\frac52}dz=\frac12 \underbrace{\oint_Cdz}_{=0}+\frac54 \oint_C\frac{1}{z-\frac52}dz$. Hence, the problem really amounts to solving $$II=\frac54\oint_C \frac{1}{z-\frac52}dz,$$ which can be done only using elementary methods. The trick in the picture below, we see that the integral along $C_1+C_2$ is equivalent to the integral along $C$ plus the integral along the clockwise circle, centered at $z=\frac52$ of radius, say, $r$. Hence, $\oint_{C_1}+\oint_{C_2}=\oint_C-\oint_{C_r}$, where $C_r$ is the positively oriented counterclockwise circle. Rearranging, we see that $$ II=\frac54 \oint_{C_1} \frac{1}{z-\frac52}dz+\frac54\oint_{C_2} \frac{1}{z-\frac52}dz+\frac54\oint_{C_r} \frac{1}{z-\frac52}dz. $$ Now the key for the first two integrals is that since we are avoiding the singularity, there exists an analytic logarithm such that $\frac{d}{dz}\log(z-\frac52)=\frac{1}{z-\frac52}$. Hence, by the fundamental theorem, the integral around a closed loop is zero. This leaves the last integral, which is easy to solve with parameterization, since the circle is centered at $z=\frac52$. enter image description here

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