∫_C (e^z /(1−cos z))dz ; C:∣z∣=1


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Question Number 115009 by arcana last updated on 22/Sep/20

$\int_{C} \frac{e^{z}}{1 - \cos z} d z; C :∣ z ∣= 1$
	Answered by Olaf last updated on 24/Sep/20

	$\int_{C} \frac{co z + i \sin z}{1 - \cos z} d z$ $\int_{C} \frac{1 + i \sin z + (\cos z - 1)}{1 - \cos z} d z$ $\int_{C} \frac{d z}{1 - \cos z} + i \int_{C} \frac{\sin z}{1 - \cos z} - \int_{C} d z$ $\int_{C} \frac{2 d z}{\sin^{2} \frac{z}{2}} + i \int_{C} \frac{\sin z}{1 - \cos z} - \int_{C} d z$ ${[- 4 \cot \frac{z}{2} + i \ln (1 - \cos z) - z]}_{C}$ $Let z = e^{i θ}, θ \in [0; 2 π]$ ${[- 4 \cot \frac{e^{i θ}}{2} + i \ln (1 - \cos e^{i θ}) - e^{i θ}]}_{0}^{2 π}$ $= 0$
	Answered by Bird last updated on 24/Sep/20
	$i give this solution but not sure I =∫_C (e^z /(1−cosz))dz we have 1−cosz ∼(z^2 /2) ⇒(e^z /(1−cosz))∼((2e^z )/z^2 ) so o is a double pole for f(z)=(e^z /(1−cosz)) I =2iπ ×Res(f,0) Res(f,o) =lim_(z→0) (1/((2−1)!)){z^2 (e^z /(1−cosz))}^((1)) =lim_(z→0) {2z^2 e^z }^((1)) =lim_(z→0) 2{2z e^z +2z^2 e^z } =0$
	$i g i v e t h i s s o l u t i o n b u t n o t s u r e$ $I = \int_{C} \frac{e^{z}}{1 - c o s z} d z w e h a v e$ $1 - c o s z \sim \frac{z^{2}}{2} \Rightarrow \frac{e^{z}}{1 - c o s z} \sim \frac{2 e^{z}}{z^{2}}$ $s o o i s a d o u b l e p o l e f o r f (z) = \frac{e^{z}}{1 - c o s z}$ $I = 2 i π \times R e s (f, 0)$ $R e s (f, o) = {l i m}_{z \to 0} \frac{1}{(2 - 1)!} {z^{2} \frac{e^{z}}{1 - c o s z}}^{(1)}$ $= {l i m}_{z \to 0} {2 z^{2} e^{z}}^{(1)}$ $= {l i m}_{z \to 0} 2 {2 z e^{z} + 2 z^{2} e^{z}}$ $= 0$
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