Power Series & Laurent Series

Solution:

Since $f^{(n)}(z) = \frac{n!}{(1-z)^{n+1}}$, we have $f^{(n)}(0) = n!$.

Therefore: $\frac{1}{1-z} = \sum_{n=0}^{\infty} \frac{n!}{n!}z^n = \sum_{n=0}^{\infty} z^n$ for $|z| < 1$.

Solution:

Using partial fractions: $\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z} = -\frac{1}{1-z} - \frac{1}{z}$

For $|z| < 1$: $-\frac{1}{1-z} = -\sum_{n=0}^{\infty} z^n$

Therefore: $f(z) = -\frac{1}{z} - \sum_{n=0}^{\infty} z^n = -\sum_{n=-1}^{\infty} z^n$

Solution:

Using Taylor series: $\sin z = z - \frac{z^3}{6} + \cdots$, so:

$\frac{\sin z}{z} = 1 - \frac{z^2}{6} + \cdots$

The Laurent series has no negative powers, so $z = 0$ is a removable singularity. We can define $f(0) = 1$ to make the function analytic.

Solution:

Since $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n$ for $|z| < 1$, differentiate:

$\frac{d}{dz}\left(\frac{1}{1-z}\right) = \frac{1}{(1-z)^2} = \sum_{n=1}^{\infty} n z^{n-1} = \sum_{n=0}^{\infty} (n+1)z^n$

So $\frac{1}{(1-z)^2} = \sum_{n=0}^{\infty} (n+1)z^n$ for $|z| < 1$.

Solution:

Using the ratio test: $\lim_{n\to\infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n\to\infty} \frac{n!}{(n+1)!} = \lim_{n\to\infty} \frac{1}{n+1} = 0$

Since the limit is 0, the radius of convergence is $R = \infty$. This is the Taylor series for $e^z$.

Solution:

Using partial fractions: $\frac{1}{z^2(z-1)} = \frac{1}{z-1} - \frac{1}{z} - \frac{1}{z^2}$

For $|z| < 1$: $\frac{1}{z-1} = -\frac{1}{1-z} = -\sum_{n=0}^{\infty} z^n$

Therefore: $f(z) = -\sum_{n=0}^{\infty} z^n - \frac{1}{z} - \frac{1}{z^2} = -\sum_{n=-2}^{\infty} z^n$

Solution:

Using the ratio test: $\lim_{n\to\infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n\to\infty} \frac{(n+1)!n^n}{(n+1)^{n+1}n!} = \lim_{n\to\infty} \frac{(n+1)n^n}{(n+1)^{n+1}}$

$= \lim_{n\to\infty} \frac{n^n}{(n+1)^n} = \lim_{n\to\infty} \frac{1}{(1+1/n)^n} = \frac{1}{e}$

Therefore, $R = e$.

Solution:

Since $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n$, differentiate twice:

$\frac{d}{dz}\left(\frac{1}{1-z}\right) = \frac{1}{(1-z)^2} = \sum_{n=1}^{\infty} n z^{n-1} = \sum_{n=0}^{\infty} (n+1)z^n$

$\frac{d}{dz}\left(\frac{1}{(1-z)^2}\right) = \frac{2}{(1-z)^3} = \sum_{n=1}^{\infty} n(n+1) z^{n-1} = \sum_{n=0}^{\infty} (n+1)(n+2)z^n$

Therefore: $\frac{1}{(1-z)^3} = \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2}z^n$

Solution:

The function $\frac{1}{1-z}$ is analytic in $\mathbb{C} \setminus \{1\}$ and equals the geometric series in $|z| < 1$.

Since the intersection is connected (the disk $|z| < 1$), and both functions are analytic there and agree, $\frac{1}{1-z}$ is the analytic continuation.

Solution:

Using partial fractions: $f(z) = \frac{1}{2}\left(\frac{1}{z-1} - \frac{1}{z+1}\right)$

For $|z| > 1$, we expand: $\frac{1}{z-1} = \frac{1}{z} \cdot \frac{1}{1-1/z} = \frac{1}{z}\sum_{n=0}^{\infty} z^{-n} = \sum_{n=1}^{\infty} z^{-n}$

Similarly: $\frac{1}{z+1} = \sum_{n=1}^{\infty} (-1)^n z^{-n}$

Therefore: $f(z) = \frac{1}{2}\sum_{n=1}^{\infty} [1 - (-1)^n]z^{-n}$

Solution:

Using $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$:

$\frac{e^z}{z^2} = \frac{1}{z^2}\sum_{n=0}^{\infty} \frac{z^n}{n!} = \sum_{n=0}^{\infty} \frac{z^{n-2}}{n!} = \sum_{n=-2}^{\infty} \frac{z^n}{(n+2)!}$

So the Laurent series has terms from $n = -2$ to $\infty$, indicating a pole of order 2 at $z = 0$.

Solution:

Using the geometric series: $\frac{1}{1+w} = \sum_{n=0}^{\infty} (-1)^n w^n$ for $|w| < 1$

Substituting $w = z^2$: $\frac{1}{1+z^2} = \sum_{n=0}^{\infty} (-1)^n z^{2n}$ for $|z| < 1$

The radius of convergence is 1 (distance to the nearest singularity at $z = \pm i$).

Solution:

$f(z) = \frac{1}{z} \cdot \frac{1}{1-z} = \frac{1}{z} \sum_{n=0}^{\infty} z^n = \sum_{n=-1}^{\infty} z^n$

This is valid for $0 < |z| < 1$.

Solution:

Using the ratio test: $\lim_{n\to\infty} \left|\frac{c_n}{c_{n+1}}\right| = \lim_{n\to\infty} \frac{2^{n+1} + 3^{n+1}}{2^n + 3^n} = \lim_{n\to\infty} \frac{3^{n+1}(2/3)^{n+1} + 1}{3^n((2/3)^n + 1)} = 3$

Therefore, $R = 3$.

Solution:

Using $\sin z = z - \frac{z^3}{6} + \frac{z^5}{120} - \cdots$:

$\frac{\sin z}{z^4} = \frac{1}{z^3} - \frac{1}{6z} + \frac{z}{120} - \cdots$

This shows a pole of order 3 at $z = 0$, with $\text{Res}(f,0) = -1/6$.

Solution:

Using the ratio test: $\lim_{n\to\infty} \left|\frac{c_n}{c_{n+1}}\right| = \lim_{n\to\infty} \frac{n+1}{n} = 1$

Therefore, $R = 1$. The series converges for $|z-1| < 1$.

Solution:

For $|z| > 1$: $\frac{1}{z-1} = \frac{1}{z}\cdot\frac{1}{1-1/z} = \sum_{n=0}^{\infty} z^{-n-1}$

For $|z| < 2$: $\frac{1}{z-2} = -\frac{1}{2}\cdot\frac{1}{1-z/2} = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$

Combining: $f(z) = \sum_{n=0}^{\infty} z^{-n-1} + \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$

Solution:

Using the geometric series: $\frac{1}{1-z^2} = \sum_{n=0}^{\infty} (z^2)^n = \sum_{n=0}^{\infty} z^{2n}$

Valid for $|z| < 1$. The radius of convergence is 1 (distance to singularities at $z = \pm 1$).

Solution:

Using partial fractions and expanding for $|z| > 1$:

$f(z) = \frac{1}{2}\left(\frac{1}{z-1} - \frac{1}{z+1}\right) = \sum_{n=0}^{\infty} z^{-n-1} - \sum_{n=0}^{\infty} (-1)^n z^{-n-1}$

Solution:

Using $\cos in = \frac{e^{-n} + e^n}{2} = \cosh n$, we have $c_n = \cosh n$.

$\lim_{n\to\infty} \left|\frac{c_n}{c_{n+1}}\right| = \lim_{n\to\infty} \frac{e^n + e^{-n}}{e^{n+1} + e^{-(n+1)}} = \lim_{n\to\infty} \frac{e^n}{e^{n+1}} = \frac{1}{e}$

Therefore, $R = 1/e$.

Solution:

This is a fundamental theorem: if $f(z) = \sum_{n=0}^{\infty} c_n(z-a)^n$ has radius of convergence $R > 0$, then $f$ is analytic in $|z-a| < R$.

Moreover, $f'(z) = \sum_{n=1}^{\infty} nc_n(z-a)^{n-1}$ and this series also has radius $R$.

Solution:

$f(z) = \frac{1}{z^2} \cdot \frac{1}{1-z} = \frac{1}{z^2}\sum_{n=0}^{\infty} z^n = \sum_{n=-2}^{\infty} z^n$

This shows a pole of order 2 at $z = 0$.

Solution:

Since $f'(z) = \frac{1}{1+z} = \sum_{n=0}^{\infty} (-1)^n z^n$ for $|z| < 1$:

$f(z) = \int_0^z f'(\zeta)d\zeta = \sum_{n=0}^{\infty} \frac{(-1)^n z^{n+1}}{n+1} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} z^n}{n}$

Valid for $|z| < 1$.

Solution:

Using partial fractions: $f(z) = \frac{1}{z-2} - \frac{1}{z-1}$

For $|z| > 1$: $\frac{1}{z-1} = \sum_{n=0}^{\infty} z^{-n-1}$

For $|z| < 2$: $\frac{1}{z-2} = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$

Combining gives the Laurent series valid for $1 < |z| < 2$.

Solution:

Using the ratio test: $\lim_{n\to\infty} \left|\frac{c_n}{c_{n+1}}\right| = \lim_{n\to\infty} \frac{(n+1)^3}{n^3} = 1$

Therefore, $R = 1$. The series converges on the closed disk $|z| \leq 1$.

Solution:

This follows from the Weierstrass M-test. For any compact subset $K$ of the disk $|z-a| < R$, there exists $r < R$ such that $K \subseteq \{z: |z-a| \leq r\}$.

Since $|c_n(z-a)^n| \leq |c_n|r^n$ and $\sum |c_n|r^n$ converges, the series converges uniformly on $K$.

Solution:

Using $\frac{1}{(1-z)^2} = \sum_{n=0}^{\infty} (n+1)z^n$:

$f(z) = \frac{1}{z^2}\sum_{n=0}^{\infty} (n+1)z^n = \sum_{n=-2}^{\infty} (n+3)z^n$

Solution:

The series $\sum_{n=0}^{\infty} z^n$ converges for $|z| < 1$ and represents $f(z) = \frac{1}{1-z}$.

The function $g(z) = \frac{1}{1-z}$ is analytic in $\mathbb{C} \setminus \{1\}$, which is a larger domain than the disk $|z| < 1$.

Therefore, $g(z) = \frac{1}{1-z}$ is the analytic continuation of $f$ to $\mathbb{C} \setminus \{1\}$.

This shows that even though the power series only converges in $|z| < 1$, the function it represents can be extended to a much larger domain.

Solution:

The series $\sum_{n=0}^{\infty} \frac{z^n}{n!}$ has radius of convergence $R = \infty$ (using the ratio test).

This series represents $e^z$, which is entire (analytic everywhere in $\mathbb{C}$).

Since the series already converges for all $z \in \mathbb{C}$, no continuation is needed—the function is already defined on the entire complex plane.

This is an example where the power series representation is valid globally, not just locally.