Evaluate-using-cauchy-s-integral-c-e-ipi-z-2-4-2-z-1-2-dz-where-c-is-a-circle-with-z-i-3-5-help-please-

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 = \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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