- #1
bobmerhebi
- 38
- 0
1. Show that f(z) = [1 - cosh(z)] / z3 has a pole as its singular point. Determine its order m & find the residue B.
2.
lim |f(z)| tends to [tex]\infty[/tex] as z tends to singular point
bm + bm+1(z-z0)+...+ b1(z-z0)m-1 + [tex]\sum[/tex][tex]^{\infty}_{n = 0}[/tex]an(z-z0)n= (z-z0) m f(z)
3. f(z) = [1 - cosh(z)] / z3 = 1/z3 - cos(iz)/z3
that's all i could do. I don't know how to do the limit of the cos part. does it approach infinity ? I think NOt right ?
please help asap.
thx in advnace
2.
lim |f(z)| tends to [tex]\infty[/tex] as z tends to singular point
bm + bm+1(z-z0)+...+ b1(z-z0)m-1 + [tex]\sum[/tex][tex]^{\infty}_{n = 0}[/tex]an(z-z0)n= (z-z0) m f(z)
3. f(z) = [1 - cosh(z)] / z3 = 1/z3 - cos(iz)/z3
that's all i could do. I don't know how to do the limit of the cos part. does it approach infinity ? I think NOt right ?
please help asap.
thx in advnace