Answer to Question #223833 in Chemical Engineering for Lokika

Question

Let f(z)=sinz/z^{4} . Then z=0 is

Answer 1

We can write

sin(z)=z-z³/6+z⁵/120-...+(-1)^kz^2k+1/(2k+1)!+...

Then

f(z)=sin(z)/z=1-z²/6+z⁴/120-...+(-1)^kz^2k/(2k+1)!+...

is Maclaurian series.

z->0 => f(z)->1

f(0)=0

So f(z) is not continuous function at z = 0 and therefore f(z) is not analytic at z = 0.

But if f(0)=1 then we have

$f(z)=\sum_{n\ge0}a_nz^n$

where

$a_n = \begin{cases} 0 &\text{if } n=2k+1 \\ (-1)^{k}/(2k)! &\text{if } n=2k \end{cases}$

$(2k)!\backsim (2k/e)^{2k}(2\pi2k)^{1/2}$

$\sqrt[2k]{(2k)!}\backsim (2k/e)(2\pi2k)^{1/{4k}}\rarr infinity$

and then

$\sqrt[n]{|a_n|}\rarr0$.

Therefore the series is covergent on the complex plane and function f(z) is analytic at z=0.

g(z)=$\int$ f(y)dy=C+z-z³/18+z⁵/600-...+(-1)^kz^2k+1/((2k+1)(2k+1)!)+...=$C+\sum_{n\ge0}b_nz^n$

where $\sqrt[n]{|b_n|}\rarr0$.

Therefore the series is covergent on complex plane and function g(z) is analytic and

$\int$^z₀f(y)dy=(C+y-y³/18+y⁵/600-...+(-1)^ky^2k+1/((2k+1)(2k+1)!)+...)|^z₀=

=z-z³/18+z⁵/600-...+(-1)^kz^2k+1/((2k+1)(2k+1)!)+...

No. There is no function f with an 

isolated singularity at 0 and such that |f(z)|~ exp( 1/|z|) near z= 0

Because in this case

$\int$_|z|=rf(z)dz does not depend on r.

But for f(z) this integral depends on r and it is equal to 2$\pi$ re^1/r