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

For the Laurent series for $\displaystyle f(z)=(z-3)\sin{\frac{1}{z+2}}$ about $\displaystyle z=-2$.
The convergence region is the disc with center in z=-2 and $\displaystyle 0<radius<5$, since we have singularities in z=3 and z=-2.
Is this correct?
The textbook unswer is that the series converges for $\displaystyle \forall z\neq -2$

The function $\displaystyle \sin\left( \frac{1}{z+2}\right)$only has only one singularity and it is at $\displaystyle z=-2$. $\displaystyle (z-3)$ is an entire function with a zero at 3.

Originally Posted by hurz

For the Laurent series for $\displaystyle f(z)=(z-3)\sin{\frac{1}{z+2}}$ about $\displaystyle z=-2$.
The convergence region is the disc with center in z=-2 and $\displaystyle 0<radius<5$, since we have singularities in z=3 and z=-2.
Is this correct?
The textbook unswer is that the series converges for $\displaystyle \forall z\neq -2$

Setting...

(1)

... You obtain...

(2)

... so that is...

(3)

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$