Evaluate using cauchy′s integral ∫_c (e^(iπ) /((z^2 +4)^2 (z+1)^2 ))dz where c is a circle with ∣z−i∣=3.5 help please


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Question Number 85260 by Umar last updated on 20/Mar/20

$Evaluate using {cauchy}^{'} s integral$ $\int_{c} \frac{e^{i π}}{{(z^{2} + 4)}^{2} {(z + 1)}^{2}} dz$ $where c is a circle with ∣ z - i ∣= 3.5$ $help please$
Commented by mathmax by abdo last updated on 20/Mar/20
$I =∫_C (e^(iπ) /((z^2 +4)^2 (z+1)^2 ))dz let W(z) =(e^(iπ) /((z^2 +4)^2 (z+1)^2 )) poles of W? W(z) =(e^(iπ) /((z−2i)^2 (z+2i)^2 (z+1)^2 )) so the poles are 2i,−2i and −1 (doubles) we have ∣2i−i∣=∣i∣=1<(7/2) ∣−2i−i∣ =∣3i∣=3<(7/2) ∣−1−i∣ =∣1+i∣=(√2)<(7/2) so ∫_C W(z)dz = 2iπ{ Res(W,2i) +Res(W,−2i) +Res(W,−1)} Res(W,2i) =lim_(z→2i) (1/((2−1)!)){ (z−2i)^2 W(z)}^((1)) =lim_(z→2i) { (e^(iπ) /((z+2i)^2 (z+1)^2 ))}^((1)) =e^(iπ) lim_(z→2i) { −((2(z+2i)(z+1)^2 +2(z+1)(z+2i)^2 )/((z+2i)^4 (z+1)^4 ))} =−e^(iπ) lim_(z→2i) {((2(2z+1)+2(z+2i))/((z+2i)^3 (z+1)^3 ))} =−e^(iπ) ×((2(4i+1)+2(4i))/((4i)^3 (2i+1)^3 )) =−e^(iπ) ×((16i +2)/(−64i (2i+1)^3 )) =e^(iπ) ×((8i+1)/(32i(2i+1)^3 )) we do the same way for Res(W,−2i) Res(W,−1) =lim_(z→−1) (1/((2−1)!)){(z+1)^2 W(z)}^((1)) =lim_(z→−1) {(e^(iπ) /((z−2i)^2 (z+2i)^2 ))}^((1)) =lim_(z→−1) e^(iπ) { (1/((z^2 +4)^2 ))}^((1)) =e^(iπ) lim_(z→−1) −((2(2z)(z^2 +4))/((z^2 +4)^4 )) =−4e^(iπ) lim_(z→−1) (z/((z^2 +4)^3 )) =4e^(iπ) ×(1/5^3 )$
$I = \int_{C} \frac{e^{i π}}{{(z^{2} + 4)}^{2} {(z + 1)}^{2}} d z l e t W (z) = \frac{e^{i π}}{{(z^{2} + 4)}^{2} {(z + 1)}^{2}} p o l e s o f W ?$ $W (z) = \frac{e^{i π}}{{(z - 2 i)}^{2} {(z + 2 i)}^{2} {(z + 1)}^{2}} s o t h e p o l e s a r e 2 i, - 2 i a n d - 1$ $(d o u b l e s) w e h a v e ∣ 2 i - i ∣=∣ i ∣= 1 < \frac{7}{2}$ $∣ - 2 i - i ∣ =∣ 3 i ∣= 3 < \frac{7}{2}$ $∣ - 1 - i ∣ =∣ 1 + i ∣= \sqrt{2} < \frac{7}{2} s o \int_{C} W (z) d z =$ $2 i π {R e s (W, 2 i) + R e s (W, - 2 i) + R e s (W, - 1)}$ $R e s (W, 2 i) = {l i m}_{z \to 2 i} \frac{1}{(2 - 1)!} {{(z - 2 i)}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to 2 i} {\frac{e^{i π}}{{(z + 2 i)}^{2} {(z + 1)}^{2}}}^{(1)}$ $= e^{i π} {l i m}_{z \to 2 i} {- \frac{2 (z + 2 i) {(z + 1)}^{2} + 2 (z + 1) {(z + 2 i)}^{2}}{{(z + 2 i)}^{4} {(z + 1)}^{4}}}$ $= - e^{i π} {l i m}_{z \to 2 i} {\frac{2 (2 z + 1) + 2 (z + 2 i)}{{(z + 2 i)}^{3} {(z + 1)}^{3}}}$ $= - e^{i π} \times \frac{2 (4 i + 1) + 2 (4 i)}{{(4 i)}^{3} {(2 i + 1)}^{3}} = - e^{i π} \times \frac{16 i + 2}{- 64 i {(2 i + 1)}^{3}} = e^{i π} \times \frac{8 i + 1}{32 i {(2 i + 1)}^{3}}$ $w e d o t h e s a m e w a y f o r R e s (W, - 2 i)$ $R e s (W, - 1) = {l i m}_{z \to - 1} \frac{1}{(2 - 1)!} {{(z + 1)}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to - 1} {\frac{e^{i π}}{{(z - 2 i)}^{2} {(z + 2 i)}^{2}}}^{(1)}$ $= {l i m}_{z \to - 1} e^{i π} {\frac{1}{{(z^{2} + 4)}^{2}}}^{(1)} = e^{i π} {l i m}_{z \to - 1} - \frac{2 (2 z) (z^{2} + 4)}{{(z^{2} + 4)}^{4}}$ $= - 4 e^{i π} {l i m}_{z \to - 1} \frac{z}{{(z^{2} + 4)}^{3}} = 4 e^{i π} \times \frac{1}{5^{3}}$
	Answered by mind is power last updated on 20/Mar/20

	$\frac{1}{2 i π} \int_{C} \frac{f (z)}{z - w} d z = f (w)$ $w e r r C i s c o n t o u r w i c h c o n t i e n w$ $w \in C$ $f^{'} (w) = \frac{1}{2 i π} \int_{C} \frac{f (z)}{{(z - w)}^{2}} d z$ $\Rightarrow \frac{1}{2 i π} \int_{C} \frac{1}{{(z - 2 i)}^{2}} \frac{e^{i π} d z}{{(z + 2 i)}^{2} {(z + 1)}^{2}} = g^{'} (2 i) + h^{'} (- 2 i) + t^{'} (- 1)$ $g (z) = \frac{e^{i π}}{{(z + 2 i)}^{2} {(z + 1)}^{2}}, h (z) = \frac{e^{i π}}{{(z - 2 i)}^{2} {(z + 1)}^{2}}, t (z) = \frac{e^{i π}}{{(z^{2} + 4)}^{2}}$ $\Rightarrow \int \frac{e^{i π} d z}{{(z^{2} + 4)}^{2} {(z + 1)}^{2}} = 2 i π (g^{'} (2 i) + h^{'} (- 2 i) + t^{'} (- 1))$ $=$
	Commented by Umar last updated on 20/Mar/20

	$thanks$
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