I'm trying to compute $$\int_\gamma \frac{z^4+z^2+1}{z^3-1} \, dz$$ where $\gamma$ is the circle $|z-i|=1$, using Cauchy's integral formula

My solution is as follows:

The integral can be rewritten as $$\int_\gamma \frac{\frac{z^4+z^2+1}{(z-1)\left(z+\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)}}{z+\frac{1}{2}-\frac{\sqrt{3}}{2}i} \, dz = I$$

Then Cauchy's integral formula can be applied, giving $$I=2\pi if\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=0$$

Is this correct?


It is almost correct. The only small problem is that, when you wrote$$\require{cancel}\frac{\ \frac{z^4+z^2+1}{(z-1)\left(z+\frac12\color{red}-\frac{\sqrt3}2i\right)}\ }{z+\frac12-\frac{\sqrt3}2i},$$you should have written$$\frac{\ \frac{z^4+z^2+1}{(z-1)\left(z+\frac12\color{red}+\frac{\sqrt{3}}{2}i\right)}\ }{z+\frac12-\frac{\sqrt3}2i}.\tag1$$But I suppose that that was a typo.

However there is an easier way of computing that integral, which avoids computing the value of the numerator of $(1)$ at $-\frac12+\frac{\sqrt2}2i$. Note that\begin{align*}\frac{z^4+z^2+1}{z^3-1}&=\frac{\frac{z^6-1}{z^2-1}}{z^3-1}\\&=\frac{\frac{z^6-1}{z^3-1}}{z^2-1}\\&=\frac{z^3+1}{z^2-1}.\end{align*}Since the distance from $\pm1$ to $i$ is greater than $1$, it follows from Cauchy's integral theorem that your integral is equal to $0$.

Note: That final rational function can be simplified further:\begin{align}\frac{z^3+1}{z^2-1}&=\frac{\cancel{(z+1)}(z^2-z+1)}{\cancel{(z+1)}(z-1)}\\&=\frac{z^2-z+1}{z-1}.\end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.