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I was trying to calculate $\int_\gamma \frac{z^2}{z-1}dz$ where $\gamma$ is a circle of radius 2, center at origin, oriented counterclockwise. I think it has a hole at z=1? I do not know how it affects the integral, what I did was using the formula $\int_\gamma f(z)dz=\int^b_af(\gamma(t))\gamma'(t)dt,$ so I got $\int_0^{2\pi}\frac{(2e^{it})^2}{2e^{it}-1}2ie^{it} dt,$ but since its starting and end points are the same, shouldn't it be zero? However the answer is not zero.

The next question is over the same contour, but a different integral $\int_\gamma \frac{e^{2z}}{(z-1)^3}dz$, and its answer is zero, I am so confused now, please help and thank you in advance

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    $\begingroup$ You need some theorems from Complex Analysis to evaluate these integrals. It is hard to get the values from just the definition. Have you heard of Cauchy's Integral Formula and Residue Theorem? $\endgroup$ Feb 20, 2021 at 7:24

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You can use Cauchy's Integral Formula: \begin{equation} f(z) = \frac{1}{2 \pi i} \int_{C_{(z_{0},r)}}\frac{f(w)}{w-z}dw \end{equation}

To use this formula $f$ must be holomorphic at an open set $\ U$ containing $\overline{D}(z_{0}, r)$ and $z \in D(z_{0}, r)$

In your case, $f(w) = w^{2}, \ z_{0} = 0, \ r = 2, \ z = 1$ so you can use the formula.

Onto your second question, the same formula can be generalized to \begin{equation} \frac{f^{n)}(z_{0})}{n!} = \frac{1}{2 \pi i} \int_{C(z_{0},r)}\frac{f(w)}{(w-z)^{n+1}}dw \end{equation} where $f^{n)}(z_{0})$ refers to the $n^{th}$ derivative.

With this formula you can solve easily both questions.

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  • $\begingroup$ are the bounds 0 and $2\pi$? $\endgroup$
    – n.y
    Feb 22, 2021 at 2:08
  • $\begingroup$ Yes, $C(z_{0},r) = z_{0} + re^{it}$ for $t \in [0, 2 \pi]$ $\endgroup$ Feb 22, 2021 at 15:08

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