complex integral with two circle (contour) against each other

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complex integral with two circle (contour) against each other

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Source: seleksi onmipa itb 2022

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#1 h Jul 27, 2022, 4:02 AM

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Let $C_1$ be a circle $|z|=3$ with counterclockwise orientation and $C_2$ be a circle $|z|=1$ with clockwise orientation.
If $f(z)=\dfrac{z^4-16z^2}{z^2+3z-10}$ , then the value of $\int_{C_1 \cup C_2} f(z) dz = \dots$

ps: i'm confused with the concept union of two contour. how i proceed? The reason behind solution is much appreciated. Thanks in advance!

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#2 h Jul 27, 2022, 4:36 AM

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The definition is $\int_{C_1} f(z) dz+\int_{C_2} f(z) dz$ , each of them can be easily calculated with the residue theorem. The only pole inside $C_1$ is $z=2$ , there are no poles inside $C_2$ .

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#3 h Jul 27, 2022, 4:48 AM

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alexheinis wrote:

The definition is $\int_{C_1} f(z) dz+\int_{C_2} f(z) dz$ , each of them can be easily calculated with the residue theorem. The only pole inside $C_1$ is $z=2$ , there are no poles inside $C_2$ .

So the integral is added up for each contour (where each contour against each other). It means, if each contour have same orientation, the integral become minus?
for the example: if $C_1$ and $C_2$ are anticlockwise, then
$\int_{C_1} f(z) dz-\int_{C_2} f(z) dz$

is my understanding true?

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#4 h Jul 27, 2022, 5:32 AM • 2 Y

Y by Mango247, pokoknyaakuimut

The orientation of $C_1$ is already contained in the name $C_1$ : it doesn't just mean the circle but the circle together with its orientation. Same for $C_2$ . So suppose $C_1=C(0,3)$ and $C_2=C(0,1)$ both anticlockwise. Then we still have $\int_{C_1\cup C_2}=\int_{C_1}+\int_{C_2}$ . Union of paths means that we run through one path and then through other the one. In each path we use the respective orientation.

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#5 h Apr 19, 2025, 12:18 PM

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The poles of $f(z) = \frac{z^4-16z^2}{z^2+3z-10} = \frac{z^4-16z^2}{(z+5)(z-2)}$ include $z = -5, 2$ . We wish to compute:

$\int_{C_{1} \cup C_{2}} f(z) dz = \int_{C_{1}} f(z) dz + \int_{C_{2}} f(z) dz$

for $C_{1}: |z|=3$ & CCW; $C_{2}:|z|=1$ & CW. Since neither pole of $f$ lies within $C_{2}$ , and only $z=2$ lies within $C_{1}$ , our computation reduces to (via Residue Theorem):

$\oint_{|z|=3} \frac{z^4-16z^2}{z^2+3z-10} dz = 2\pi i \cdot Res_{z\rightarrow 2}f(z) = 2\pi i\left[\frac{z^4-16z^2}{2z+3}\right]_{z=2} = \textcolor{red}{-\frac{96\pi i}{7}}$ .

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