# Evaluating $\int_{C}\left({{e^{2z}\over z^2(z^2+2z+2)}+\ln(z-6)+{1\over (z-4)^2 }}\right) dz$.

Question : Evaluate

$$\int_{C}\left({{e^{2z}\over z^2(z^2+2z+2)}+\log(z-6)+{1\over (z-4)^2 }}\right) dz$$

where C is the circle $|z|=3$. State the theorems your have used to evaluate the integral

My Attempt : Integral of the third term becomes $0$ as 4 lies outside the given circle so is analytic everywhere and so by cauchy's theorem it must be 0

The first integral has pole of order 2 at 0 and another complex pole of order 1 for $z^2+2z+2$ which can be solved using cauchy's formula for residues.

My problem is with the middle integral $\log(z-6)$ as we know that $\log(z-6)$ will not be analytic in the circle $|z|=3$ as it attains negative values which is not defined. so how to go about this integral.

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@Mhenni Benghorbal Thanks for your kind reply !! I mean the second term of the integral :) .. Can you please elaborate how to ork around the second integral with log ? i have this idea that points inside the circle $|Z|=3$ will yeild negative values for log which are not defined and hence $log(z-6)$ is not analytic in the circle. I know i am missing out on some fundamental concept here. Help is greatly appreciated.Thanks – Aman Mittal Apr 25 '13 at 7:16
I just deleted my hint because you had already got your poles. – Mhenni Benghorbal Apr 25 '13 at 7:18
What do you think $\ln(-1)$? – Mhenni Benghorbal Apr 25 '13 at 7:19
@AmanMittal: Maybe you could take a different branch of $\log$? What is their definition of $\log$? – wj32 Apr 25 '13 at 7:20
can you please elaborate on integral for log(z-6) to explain how you got 0 ? i am not able to see why log(z-6) will be analytic in the given region. I am probably missing out on some concept or have a wrong one :( – Aman Mittal Apr 25 '13 at 7:20

Recall that a function logarithm can be defined on each domain of the complex plane with no loop around zero. For example, the usual (natural) logarithm $\ln$ is often first defined on $\mathbb R_+^*$ and it can be extended to $\mathbb C\setminus \{z\in\mathbb C\mid\Re(z)\leqslant0,\Im(z)=0\}$ in the way you know. But a logarithm function $\mathrm{Log}$ also exists on $\mathbb C\setminus D$ where $D=\{z\in\mathbb C\mid\Re(z)\geqslant0,\Im(z)=0\}$, which may be defined as follows: each $z$ in $\mathbb C\setminus D$ can be written uniquely as $z=r\mathrm e^{\mathrm it}$ with $r$ and $t$ real, $r\gt0$, $t$ in $(0,2\pi)$, then $\mathrm{Log}(z)=\ln(r)+\mathrm it$ (other, more intrinsic, definitions exist but this one will do).

This $\mathrm{Log}$ function (or one of its close analogs) seems to be what is meant in your question. Then $z\mapsto\mathrm{Log}(z-6)$ is holomorphic on $|z|\lt6$ and the circle $C=\{z\in\mathbb C\mid|z|=3\}$ is included in this domain hence $$\oint_C\mathrm{Log}(z-6)\mathrm dz=0.$$

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can i do this : \\ Root of $z-6=0$ is 6 so 6 is the branch point of $log(z-6)$ and hence a singularity . Now distance of the point $z=0 \from z=6$ is 6, hence, $log(z-6)$ is analytic in $|Z|<6$ and so analytic in $|z|=3$. And hence the value of integral is 0 as said. Thanks :) – Aman Mittal Apr 25 '13 at 8:23

Hint: You have three poles within your contour, namely $z=0$ with order $2$, $z=-1-i$, and $z=-1+i$. Now, all you need to do is to find the residues at every pole and add them. See here for a formula for evaluating residues

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yes, that is true only for the first integral. what about the second integral log(z-6) – Aman Mittal Apr 25 '13 at 6:55
@AmanMittal: You have only one integral! – Mhenni Benghorbal Apr 25 '13 at 7:04
@AmanMittal: Note that, $\ln(z-6)$ has a branch point at $z=6$ which is out side the contour. However you can split your integral to three integrals and the last two integrals have value zero, since their integrands are analytic functions for the contour you have been given. – Mhenni Benghorbal Apr 25 '13 at 7:07