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# Answer to Question #223833 in Chemical Engineering for Lokika

Question #223833

Let f(z)=sinz/z^{4} . Then z=0 is

1
2021-08-18T09:08:02-0400

We can write

sin(z)=z-z3/6+z5/120-...+(-1)kz2k+1/(2k+1)!+...

Then

f(z)=sin(z)/z=1-z2/6+z4/120-...+(-1)kz2k/(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-z3/18+z5/600-...+(-1)kz2k+1/((2k+1)(2k+1)!)+...=

where .

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

z0f(y)dy=(C+y-y3/18+y5/600-...+(-1)ky2k+1/((2k+1)(2k+1)!)+...)|z0=

=z-z3/18+z5/600-...+(-1)kz2k+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|=rf(z)dz does not depend on r.

But for f(z) this integral depends on r and it is equal to 2 re1/r

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