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

**hurz** · April 28th 2011, 03:18 PM

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 $0<radius<5$ , since we have singularities in z=3 and z=-2.
Is this correct?
The textbook unswer is that the series converges for $\forall z\neq -2$

**TheEmptySet** · April 28th 2011, 04:35 PM

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.

**chisigma** · April 29th 2011, 01:16 AM

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 $0<radius<5$ , 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

$\chi$ $\sigma$

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

LinkBack

Thread Tools

Display

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

Search Tags