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

1. ## 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 $\forall z\neq -2$

2. 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.

3. 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...

(1)

... You obtain...

(2)

... so that is...

(3)

The (3) converges everywhere is and has at z=-2 an essential singularity...

Kind regards

$\chi$ $\sigma$