Complex Integration using Cauchy's Theorem

Question

The problem is the integration of $$I=\int_{\left\lvert z-1\right\rvert=1} f(z) dz$$ where $$f(z)=\frac{1}{z^3-1}$$ and the path goes $1$ loop in positive direction.

I tried to solve the problem using Cauchy's Theorem by finding $z$ that makes $f(z)$ denominator be $0$. That was $z=1$.

And I got struck.

I think the integral is needed to be treat somehow so that

$$\int_{C_a}\frac{1}{(z-\alpha)^n}dz = 2{\pi}i \text{ when } n=1$$

can be used.

My question is how should I continue with the integral?

Answer 1

The only pole of $f(z)$ on the circle is $z=1$

According to residue theorem $$\oint _{\gamma }f(z)\,dz=2\pi i \operatorname {Res} (f,1)$$ And we have $\operatorname {Res} (f,1)=\lim_{z\to 1} \, \dfrac{z-1}{z^3-1}=\lim_{z\to 1} \, \dfrac{1}{z^2+z+1}=\dfrac{1}{3}$

Thus $\oint _{\gamma }f(z)\,dz=\dfrac{2 i \pi }{3}$

Hope this helps

Answer 2

Hint: From $z^3-1=(z-1)(z^2+z+1)$ look at the following function $g(z)=\frac{1}{z^2+z+1}$. Then from Cauchy's integral formula, since $1 \in \{ z : | z − 1| < 1\}$ $$g(1)=\frac{1}{2\pi i}\int_{|z−1|=1} \frac{g(z)}{z-1}dz=\frac{1}{2\pi i}\int_{|z−1|=1} \frac{1}{z^3-1}dz$$

Answer 3

Determine which roots of the denominator are contained within the given curve $|z - 1| = 1$, find the residues at those roots, and calculate using the Residue Theorem.

Answer 4

Note: The hint of @rtybase is essentially the answer to OPs question. Let's recall Cauchy's Integralformula:

If $a$ is in the interior of $\gamma=\{z:|z-1|=1\}$ and a function $g$ is holomorphic in a region which contains the closure of the interior of $\gamma$, then \begin{align*} g(a)=\frac{1}{2\pi i}\oint_{\gamma} \frac{g(z)}{z-a}\,dz \end{align*}

We have the following situation:

The function $f(z)=\frac{1}{z^3-1}$ has three simple poles at $z_0=1$ and $z_{1,2}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$. The pole $z_0=1$ is in the interior of $\gamma$. We can write $f(z)$ as \begin{align*} f(z)=\frac{1}{z^3-1}=\frac{1}{(z-1)(z^2+z+1)} \end{align*} and observe: Since $g(z)=\frac{1}{z^2+z+1}$ is holomorph in the interior of $\gamma$ we obtain \begin{align*} g(1)&=\frac{1}{2\pi i}\oint_{\gamma} \frac{g(z)}{z-1}\,dz\\ &=\frac{1}{2\pi i}\oint_{\gamma} \frac{1}{(z-1)(z^2+z+1)}\,dz\\ &=\frac{1}{2\pi i}\oint_{\gamma} f(z)\,dz\\ \end{align*}

We finally conclude \begin{align*} \oint_{\gamma} f(z)\,dz=2\pi i\, g(1)=\frac{2\pi i}{3} \end{align*}