Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can't find out where did I go wrong in solving:

enter image description here

$\displaystyle\int_{|z|=1}\frac{4+z}{(2-z)z}\\=\displaystyle\int_{|z|=1}\frac{3z+2(2-z)}{(2-z)z}\\=-3\displaystyle\int_{|z|=1}\frac{dz}{z-2}+2\displaystyle\int_{|z|=1}\frac{dz}{z}\\=0+2\text{Res$_{z=0}\dfrac{1}{z}$}\text{ (by Cauchy's Integral Formula)}\\=4\pi i$

share|cite|improve this question
You've missed off several $\operatorname{d}\!z$s. – Fly by Night Aug 6 '13 at 15:08
$$\int_{|z|=1}\frac{3z+2(2-z)}{(2-z)z}\,dz=-3\int_{|z|=1}\frac{dz}{z-2}+2\int_{|‌​z|=1}\frac{dz}{z}.$$ – Adrian Keister Aug 6 '13 at 15:09
You mutated a $z-2$ into a $z-1$, and $\int_{\lvert z\rvert = 1} \frac{dz}{z-1}$ doesn't converge. – Daniel Fischer Aug 6 '13 at 15:09
After the edit, the first integral is wrong. (The integrand is analytic on the unit disc.) – mrf Aug 6 '13 at 15:12
Note that $2$ is not inside the unit circle, so that $\int \frac{dz}{z-2} = 0$. – Thomas Andrews Aug 6 '13 at 15:18
up vote 0 down vote accepted

After you break it down to two integrals, take another look at your first integral. The pole for that first inegral is at $z=2$, but your path is the circle defined by $|z|=1$. The residue theorem gives a non-zero result only if the pole is inside the closed contour of the integral. The pole at $z=2$ is not inside your path, so it will give zero to your integral, and not $-3\times 2\pi i$.

share|cite|improve this answer

You have an algebraic mistake in the partial fractions decomposition. After the edit, the first integral is wrong. (The integrand is holomorphic on the unit disc.)

A simpler aproach:

$$\int_{|z|=1} \frac{\frac{4+z}{2-z}}{z}\,dz = 2\pi i \cdot \left. \frac{4+z}{2-z} \right|_{z=0} = 4\pi i$$ by Cauchy's integral formula. Note that $$f(z) = \frac{4+z}{2-z}$$ is holomorphic on (a neighboorhood) of the unit disc.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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