Question-160596

Question Number 160596 by Ahmed1hamouda last updated on 03/Dec/21

Answered by Mathspace last updated on 03/Dec/21

I = \int_{c} \frac{e^{3 z} s i n (2 z)}{{(4 z - j π)}^{3}} d z w i t h c \to∣ z - j ∣= 1

l e t φ (z) = \frac{e^{3 z} s i n (2 z)}{{(4 z - j π)}^{3}} = \frac{e^{3 z} s i n (2 z)}{64 {(z - \frac{j π}{4})}^{3}}

z = \frac{j π}{4} (i t a k e j = e^{\frac{i 2 π}{3}})

∣ \frac{j π}{4} - j ∣=∣ \frac{π}{4} - 1 ∣< 1 \Rightarrow

I = 2 i π R e s (φ, \frac{j π}{4})

R e s (φ, \frac{j π}{4}) = {l i m}_{z \to \frac{j π}{4}} \frac{1}{(3 - 1)!} {{(z - \frac{j π}{4})}^{3} φ (z)}^{(2)}

= {l i m}_{z \to \frac{j π}{4}} \frac{1}{128} {e^{3 z} s i n (2 z)}^{(2)}

= \frac{1}{128} {l i m}_{z \to \frac{j π}{4}} {3 e^{3 z} s i n (2 z) + 2 e^{3 z} c o s (2 z)}^{(1)}

= \frac{1}{128} {l i m}_{z \to \frac{j π}{4}} (9 e^{3 z} s i n (2 z) + 6 e^{3 z} c o s (2 z) + 6 e^{3 z} c o s (2 z) - 4 e^{3 z} s i n (2 z))

= \frac{1}{128} {9 e^{\frac{3 j π}{4}} s i n (\frac{j π}{2}) + 12 e^{\frac{3 j π}{4}} c o s (\frac{j π}{2}) - 4 e^{\frac{3 j π}{4}} s i n (\frac{j π}{2}))

Commented by Ahmed1hamouda last updated on 03/Dec/21

In which reference engineering mathematics is there this issue

Commented by Mathspace last updated on 03/Dec/21

c o m p l e x a n a l y s i s a n d t h e o r e m o f r e s i d u s

Commented by Ahmed1hamouda last updated on 03/Dec/21

What is the name of the author of this reference?