Find the Laurent series for the following functions about 0 and classify their singularities there.
\(\frac { \sin ^ { 2 } z } { z }\)
\(\scriptstyle z \exp ( { \frac { 1 } { z ^ { 2 } } } )\)
\(\frac { 1 } { z ( 4 - z ) }\)
Solution
1. For part (a),\[ \boxed{ \frac{\sin^2 z}{z} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}2^{2n-1}}{(2n)!}z^{2n-1}. } \]The singularity at \(0\) is removable.
Using \[ \sin^2z=\frac{1-\cos(2z)}2 \] and the Taylor series for cosine, \[ \cos(2z) = \sum_{n=0}^{\infty} \frac{(-1)^n(2z)^{2n}}{(2n)!}, \] one gets \[ 1-\cos(2z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}2^{2n}z^{2n}}{(2n)!}. \] Dividing by \(2z\) gives the displayed series. It contains no negative powers, so the singularity at \(0\) is removable. The series has infinite radius of convergence and represents the holomorphic extension at \(0\).
2. For part (b),\[ \boxed{ z e^{1/z^2} = \sum_{n=0}^{\infty}\frac{z^{1-2n}}{n!}. } \]The singularity at \(0\) is essential.
The exponential series gives \[ e^{1/z^2} = \sum_{n=0}^{\infty}\frac{z^{-2n}}{n!} \] for every \(z\neq0\). Multiplying by \(z\) yields the displayed Laurent series, valid on \[ 0<\abs{z}<\infty. \] Its principal part contains infinitely many nonzero terms, \[ z^{-1}+\frac{z^{-3}}{2!}+\frac{z^{-5}}{3!}+\cdots. \] Therefore \(0\) is an essential singularity.
3. For part (c), on the annulus \(0<\abs{z}<4\),\[ \boxed{ \frac1{z(4-z)} = \sum_{n=0}^{\infty}\frac{z^{n-1}}{4^{n+1}}. } \]In particular, the singularity at \(0\) is a simple pole.
For \(\abs{z}<4\), \[ \begin{aligned} \frac1{z(4-z)} &= \frac1{4z}\frac1{1-z/4}\\ &= \frac1{4z} \sum_{n=0}^{\infty}\left(\frac z4\right)^n\\ &= \sum_{n=0}^{\infty}\frac{z^{n-1}}{4^{n+1}}. \end{aligned} \] The principal part consists of the single nonzero term \[ \frac1{4z}. \] Hence \(0\) is a pole of order \(1\).
4. For part (c), on the other maximal annulus \(\abs{z}>4\),\[ \boxed{ \frac1{z(4-z)} = -\sum_{n=0}^{\infty}4^n z^{-n-2}. } \]
If \(\abs{z}>4\), then \[ \begin{aligned} \frac1{z(4-z)} &= -\frac1{z^2}\frac1{1-4/z}\\ &= -\frac1{z^2} \sum_{n=0}^{\infty}\left(\frac4z\right)^n\\ &= -\sum_{n=0}^{\infty}4^n z^{-n-2}. \end{aligned} \] The two annuli in steps 3 and 4 are maximal because the other finite singularity is at \(z=4\).
5. The singularities at \(0\) are respectively removable, essential, anda simple pole.
Step 1 gives a Laurent series with no principal part; step 2 gives infinitely many negative-power terms; and step 3 gives exactly one negative-power term of order \(1\). These are precisely the three stated classifications.
6. Q.E.D.
Steps <1>1–<1>5 provide the requested Laurent series and classifications.