# contour integral computations

Let $C$ be the boundary of the square whose vertices are $1+i$, $1-i$, $-1 + i$ and $-1 -i$. Suppose that $C$ is oriented counterclockwise. How to compute

a) $$\int_C \frac{e^z}{z-1/2} \, dz$$

b) $$\int_C \ln(z+3) \, dz$$

c) $$\int_C \bar{z} \, dz$$

I seee that we can use $f'(z)z'$ and maybe we can proceed this using a definition of Cauchy Riemann?

Also, can we use the fundamental theorem of calculus on this?

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You may find the anti-derivative to evaluate this integral. What have you tried so far? If this is homework, you may want to consider adding the "homework" tag. –  Argon Jul 13 '12 at 19:57
Do you know the residue theorem? –  Fabian Jul 13 '12 at 20:01
I've heard about it, but thats for a factor of 2pi*i –  mary Jul 13 '12 at 20:07
Try parametrizing the contour. –  Argon Jul 13 '12 at 20:08
@Argon : I think what you suggest is needlessly complicated. What Fabian suggests solves the problem much more simply. –  Michael Hardy Jul 13 '12 at 20:54

In your first integral, the numerator is an entire function that has no zeros. Since it's an entire function, it has no poles, so the fraction can have a pole only if the denominator has a zero. Since the numerator has no zeros, every place where the denominator has a zero will be a pole of the function defined by the fraction. The point where the denominator has a zero is INSIDE the square. The value of the numerator at that point is $e^{1/2}$. So the integral is equal to $$e^{1/2}\int_C \frac{dz}{z-1/2}.$$ In general, the integral $\displaystyle\int_B \frac{dz}{z-a}$ depends only on whether $a$ is inside or outside of the curve $B$, if $B$ winds once around every point that it surrounds.

Is that enough for you to figure out the first one? (If not, I can add more.)

The second one has a branch point at $-3$ but behaves well within the square you've describe, i.e. it's holomorphic on the curve and in the region it winds around. That gets you the value of the integral almost instantly, by citing a well known theorem.

For the third one, the integral of the complex conjugate of a function is just the complex conjugate of the integral. So find $\displaystyle \int_C z~dz$ by citing the theorem mentioned in the paragraph above, and take its conjugate.

Later note: @mary : Your comment above about the residue theorem suggests you might have some difficulties with what I wrote above. The $2\pi i$ that you mention is indeed the value of the integral $\displaystyle\int_C\frac{dz}{z-1/2}$, and if you're arare of that "$2\pi i$", you can probably figure that out. However, the residue theorem is not limit to finding integrals whose value is $2\pi i$. Rather, the residue theorem says that the integral along a curve that winds once around all the points it surrounds is $2\pi i$ times the sum of the residues at all of the points surrounded by the curve where a singularity occurs. (At least if one assumes only finitely many such points, and that's enough for the present occasion.) Where you integrate $z~dz$, you're integrating a function that has no singularities, hence that sum is $0$. When you integrate $\ln(z+3)~dz$, there are no sigularities at points that the curve winds around, so again that sum is $0$.

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For reference, the first concept Michael is trying to explain is called the Cauchy integral formula which says if $z_0$ is inside our contour $C$ and $f$ is analytic on and within $C$ then $$f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}$$ –  Ahsan Jul 13 '12 at 21:02
-1, I don't believe "the integral of the complex conjugate of a function is just the complex conjugate of the integral" is true. See this mse question. –  orlandpm Mar 24 '13 at 1:00
@orlandpm : You're right, and I thought I'd corrected that some time ago. Now I'm wondering it that was in some other thread. –  Michael Hardy Mar 24 '13 at 1:43

For c) $\int_C \overline{z} dz=2i {\rm Area} (R)$ where $R$ is the interior of $C$. This follows from Green's theorem since $\int_C (x-iy)(dx+idy)=\int_C xdx-ydy +i \int_C -ydx+xdy =2i \int \int_{R} dx dy =2i {\rm Area} \ (R)$

For a) Use Cauchy integral formula

$f(a)=\frac{1}{2\pi i} \int_{C} \frac{f(z)}{z-a} \ dz$

with $f(z)=e^z$ and $a=\frac{1}{2}$.

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It is equal to $2\pi i e^{1/2}$, by the Cauchy integral theorem. Notice that the curve $C$ winds about the point $z = 1/2$ exactly once.

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