# f(z)=(z-3)sein(1/(2+z))

• Apr 28th 2011, 02:18 PM
hurz
f(z)=(z-3)sein(1/(2+z))
For the Laurent series for $f(z)=(z-3)\sin{\frac{1}{z+2}}$ about $z=-2$.
The convergence region is the disc with center in z=-2 and $0, since we have singularities in z=3 and z=-2.
Is this correct?
The textbook unswer is that the series converges for $\forall z\neq -2$
• Apr 28th 2011, 03:35 PM
TheEmptySet
The function $\sin\left( \frac{1}{z+2}\right)$only has only one singularity and it is at $z=-2$. $(z-3)$ is an entire function with a zero at 3.
• Apr 29th 2011, 12:16 AM
chisigma
Quote:

Originally Posted by hurz
For the Laurent series for $f(z)=(z-3)\sin{\frac{1}{z+2}}$ about $z=-2$.
The convergence region is the disc with center in z=-2 and $0, since we have singularities in z=3 and z=-2.
Is this correct?
The textbook unswer is that the series converges for $\forall z\neq -2$

Setting...

http://quicklatex.com/cache3/ql_86a4...0fa22b0_l3.png (1)

... You obtain...

http://quicklatex.com/cache3/ql_1a32...b2078b6_l3.png (2)

... so that is...

http://quicklatex.com/cache3/ql_493b...45b24b1_l3.png (3)

The (3) converges everywhere is http://quicklatex.com/cache3/ql_0571...c037d33_l3.png and has at z=-2 an essential singularity...

Kind regards

$\chi$ $\sigma$