Question #223833

Let f(z)=sinz/z^{4} . Then z=0 is

Expert's answer

We can write

sin(z)=z-z^{3}/6+z^{5}/120-...+(-1)^{k}z^{2k+1}/(2k+1)!+...

Then

f(z)=sin(z)/z=1-z^{2}/6+z^{4}/120-...+(-1)^{k}z^{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

where

and then

.

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

g(z)= f(y)dy=C+z-z^{3}/18+z^{5}/600-...+(-1)^{k}z^{2k+1}/((2k+1)(2k+1)!)+...=

where .

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

^{z}_{0}f(y)dy=(C+y-y^{3}/18+y^{5}/600-...+(-1)^{k}y^{2k+1}/((2k+1)(2k+1)!)+...)|^{z}_{0}=

=z-z^{3}/18+z^{5}/600-...+(-1)^{k}z^{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

_{|z|=r}f(z)dz does not depend on r.

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

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