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How would you integrate to show $$\int_{\partial B(1,2)} \frac{1}{(z-2)^3}dz = 0$$ where $B(1,2)$ is a ball centre 1, radius 2 in the complex plane. Apparently you're meant to use the FTC but I can't see how to use the FTC in the complex plane?


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Is FTC the Fundamental theorem of calculus? – draks ... Apr 10 '12 at 17:13
Yeah. This question is on a past paper and the answers are really brief and literally say 'FTC' above the equals sign. Oh and this integral is $0$ by the way, I'll edit that in. – user26069 Apr 10 '12 at 17:16
Follows from Cauchys Integral theorem or some of the more primitive versions of it (which might have been meant by "FTC"?). Wikipedia also compares it to the fundamental theorem of calculus: – example Apr 10 '12 at 17:23
Is there any possibility that you intended $(z-2)^3$ instead of $(z-2^3)$? – Michael Hardy Apr 10 '12 at 17:50

First way: Using Cauchy's differentiation formula $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_\gamma\frac{f(z)}{(z-a)^{n+1}}\,dz$$ we get that$$\oint_{|z-1|=2}\frac{1}{(z-2)^3}\,dz=\left.\frac{2\pi i}{2!}\,\frac{\mathcal d^2}{\mathcal d z^2}(1)\,\,\right|_{z=2}=0$$

Second way: Making first the translation $\,z\to z+1\,$ , we take the circle of radius $\,2\,$ centered at the origin and $\,(z-1)^3\,$ in the denominator, getting $$I:=\oint_{|z|=2}\frac{1}{(z-1)^3}\,dz$$

Now, we put $\,z=2e^{i\theta}\,\,,\,0\leq\theta\leq 2\pi\Longrightarrow dz=2ie^{i\theta}\,$ , so we get$$I=\int_0^{2\pi}\frac{2ie^{i\theta}d\theta}{\left(2e^{i\theta}-1\right)^3}=\int_0^{2\pi}\frac{(2e^{i\theta})}{\left(2e^{i\theta}-1\right)^3}=\left.-\frac{1}{2}\frac{1}{(2e^{i\theta}-1)^2}\right|_0^{2\pi}=$$$$=-\frac{1}{2}\left(\frac{1}{(2-1)^2}-\frac{1}{(2-1)^2}\right)=0$$

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We are told to compute the given integral $Q$ without the benefits of the Cauchy theorem. The circle in question can be parametrized as $$\gamma:\quad \phi\mapsto 1+2e^{i\phi}\qquad(0\leq\phi\leq2\pi)\ ,$$ which gives $dz=2i\,e^{i\phi}\,d\phi$. Therefore we have $$Q=\int_0^{2\pi}{2i e^{i\phi}\over \bigl(2e^{i\phi}-1\bigr)^3}\ d\phi\ .\qquad(*)$$ Note that the denominator in the last integral is $\ne0$ throughout.

The function ${1\over(z-2)^3}$ appearing in the definition of $Q$ has a primitive, namely $-{1\over 2(z-2)^2}$. As a consequence the complex function of the real variable $\phi$ appearing in $(*)$ also has a simple primitive. In fact it is easily checked that $$Q=-{1\over2\bigl(2e^{i\phi}-1\bigr)^2}\Biggr|_0^{2\pi}\ ,$$ and since $e^{i\phi}$ assumes the same value at $0$ and at $2\pi$ the difference on the right hand side is $0$.

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FTC way: $$ \frac{1}{(z-2)^3} = \frac{d}{dz} \frac{-1}{2(z-2)^2} $$ everywhere on your curve, so to get the value of the integral along the curve, evaluate $-1/(2(z-2)^2))$ at the two endpoints and subtract. Of course if it is a closed curve, that difference is zero.

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No one has mentioned residues, so I will throw that in, too.

Let $w=z-2$, and the integral becomes $$ \oint_{\partial B(-1,2)}\frac{1}{w^3}\,\mathrm{d}w $$ The coefficient of the $\dfrac1w$ term in the Laurent series is $0$, so the residue at the singularity at $w=0$ is $0$ even though the path of integration, $|w+1|=2$, circles it. Thus, $$ \oint_{\partial B(1,2)}\frac{1}{(z-2)^3}\,\mathrm{d}z=0 $$

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