let-f-z-sin-2z-z-n-with-n-integr-natural-calculate-Res-f-0-

Question Number 63079 by mathmax by abdo last updated on 28/Jun/19

l e t f (z) = \frac{s i n (2 z)}{z^{n}} w i t h n i n t e g r n a t u r a l

c a l c u l a t e R e s (f, 0)

Commented by mathmax by abdo last updated on 03/Jul/19

0 i s p o l e o f f a t o r d r e n \Rightarrow R e s (f, 0) = {l i m}_{z \to 0} \frac{1}{(n - 1)!} {z^{n} f (z)}^{(n - 1)}

= {l i m}_{z \to 0} \frac{1}{(n - 1)!} {s i n (2 z)}^{(n - 1)}

w e h a v e {s i n}^{(1)} (2 z) = 2 c o s (2 z) = 2 s i n (2 z + \frac{π}{2}) l e t s u p p o s e t h a t

{s i n}^{(n)} (2 z) = 2^{n} s i n (2 z + \frac{n π}{2}) \Rightarrow {s i n}^{(n + 1)} (2 z) = 2^{n + 1} c o s (2 z + \frac{n π}{2})

= 2^{n + 1} s i n (2 z + (n + 1) \frac{π}{2}) s o t h e r e l a t i o n i s t r u e a t o r d r n + 1 \Rightarrow

R e s (f, 0) = {l i m}_{z \to 0} \frac{1}{(n - 1)!} \times 2^{n - 1} s i n (2 z + \frac{(n - 1) π}{2})

= \frac{2^{n - 1}}{(n - 1)!} s i n (\frac{(n - 1) π}{2}) w i t h n ⩾ 1 .