Calculate the integral ∮dzez−1 over the circle of radius r=4 centered at z=4i

We have to evalute dzez−1 over the circle of radius r=4 centered at z=4i the point z=2πi lies at the circle at the point ez−1 becomes zero. Therefore f(z)=1ez−1 has a simple pole at z=2πi then by the cauchy integral formula ∫1ez−1 residue at z=2πi. Now from the difinition residue at z=2πi is limN→2πi(z−2πi) f=limN→2πi(z−2πi)1ez−1=1e2πi=1 Thus, ∫1ez−1=2πi

Expert answers to "Calculate the integral ∮dzez−1 over the circle of..."

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Calculus and Analysis
Integral Calculus
Integrals

FizeauV

Answered question

2021-08-12

Calculate the integral

\oint \frac{d z}{e^{z - 1}}

over the circle of radius r=4 centered at z=4i

Answer & Explanation

opsadnojD

Skilled2021-08-13Added 95 answers

We have to evalute $\frac{d z}{e^{z - 1}}$ over the circle of radius $r = 4$ centered at $z = 4 i$ the point $z = 2 π i$ lies at the circle at the point $e^{z - 1}$ becomes zero.
Therefore $f (z) = \frac{1}{e^{z} - 1}$ has a simple pole at $z = 2 π i$ then by the cauchy integral formula $\int \frac{1}{e^{z} - 1}$ residue at $z = 2 π i$ .
Now from the difinition residue at $z = 2 π i$ is $lim_{N \to 2 π i} (z - 2 π i)$

$f = lim_{N \to 2 π i} (z - 2 π i) \frac{1}{e^{z} - 1} = \frac{1}{e^{2} π i} = 1$
Thus, $\int \frac{1}{e^{z} - 1} = 2 π i$

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