Question

Let f(z) be an analytic function in the annulus 0 <|z| < R for some positive real number R,Whose laurent series (in this annulus) is given by

f(z) = n from -∞ to ∞ ∑ { (-1)^n / (n^2)! ] } . Z ^ { 5n - n^2 -1}
A)) What Kind of Singularity is z=0 for f(z) ?
B)) Compute integral on Curve for [ z ^ 24 . f(z) dz] ,where C is a counterclockwise simple path lying in the annulus enclosing z=0
C)) Calculate Res (f) in z=0
D)) Evaluate Integral on Curve for [ sin Z .f(z) dz] , where C : |z| = (R/2) oriented counterclockwise
Note : please not by limit

Answer 1

Answer on Question #50231- Math – Complex Analysis

Let $f(z)$ be an analytic function in the annulus $0 < |z| < R$ for some positive real number $R$. Whose laurent series (in this annulus) is given by

$$f(z) = n \text{ from } -\infty \text{ to } \infty \sum \left\{ (-1)^n / (n^2)! \right\} \quad \text{. Z}^{\wedge} \{5n - n^2 - 1\}$$

A) What kind of singularity is $z=0$ for $f(z)$?

B) Compute integral on Curve for $[z^{\wedge} 24 \cdot f(z) \, dz]$, where $C$ is a counterclockwise simple path lying in the annulus enclosing $z=0$.

C) Calculate Res (f) in $z=0$.

D) Evaluate Integral on Curve for $[\sin Z \cdot f(z) \, dz]$, where $C: |z| = (R/2)$ oriented counterclockwise.

Solution

A) The function

$$f(z) = \sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{(n^2)!} z^{5n - n^2 - 1} = \sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{(n^2)!} \frac{1}{z^{n^2 - 5n + 1}}$$

has an singularity at point $z_0 = 0$, because it is not defined there, but it is defined at other points of the annulus $0 < |z| < R$.

The function $f(z)$ has an essential singularity at 0, because according to (1), $f(z)$ contains infinitely many terms with negative powers of $z$ ($5n - n^2 - 1 < -1$ for $n < 0$ and $n > 5$), i.e. there exist infinitely many terms with negative power of $z$.

B) The function

$$z^{24}f(z) = z^{24} \sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{(n^2)!} z^{5n - n^2 - 1} = \sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{(n^2)!} \frac{1}{z^{n^2 - 5n - 24 + 1}}$$

has an essential singularity at point $z_0 = 0$.

We used $z^{24}z^{5n - n^2 - 1} = z^{24 + 5n - n^2 - 1} = z^{-(n^2 - 5n - 24 + 1)} = \frac{1}{z^{n^2 - 5n - 24 + 1}}$ according to properties of exponents and power functions, operations with real numbers.

Coefficient $c_{-1}$ of the term with $z^{-1} = \frac{1}{z} = \frac{1}{z^1}$ in (2) is called the residue of function $h(z) = z^{24}f(z)$ at point $z_0 = 0$, here point $z_0 = 0$ is finite.

(notation is the following: $Res_{z=0}(h) = Res_{z=0}(z^{24}f) = c_{-1}$).

In order to find the coefficient that corresponds to values of $n$ such that

$$\frac{1}{z^{n^2 - 5n - 24 + 1}} = \frac{1}{z^1}$$

, search for natural solutions to equation

$$n^2 - 5n - 24 + 1 = 1,$$

which is equivalent to quadratic equation.

$$n^{2} - 5n - 24 = 0,$$

its discriminant is

$$D = (-5)^{2} - 4 \cdot (-24) = 25 + 4 \cdot 24 = 25 + 96 = 121,$$ $$\text{hence } n = \frac{5 \pm 11}{2} = \frac{5 - 11}{2}; \frac{5 + 11}{2} = \frac{-6}{2}; \frac{16}{2} = -3; 8$$

which implies that we take only natural solution $n = 8$, therefore, the coefficient $c_{-1}$ of the term of (2) with $z^{-1}$ will be $\frac{(-1)^n}{(n^2)!} = \frac{(-1)^8}{(8^2)!} = \frac{1}{64!}$, which leads to

$$Res_{z=0}(h) = Res_{z=0}(z^{24}f) = c_{-1} = \frac{1}{64!}.$$

Coefficient $c_{-1}$ can be computed as $c_{-1} = \frac{1}{2\pi i}\oint_T\frac{h(z)}{(z - 0)^{-1 + 1}} = \frac{1}{2\pi i}\oint_Th(z)dz$, here $T$ is a path lying in the annulus $(0 < |z| < R)$ enclosing $z = 0$.

Thus, by residue theorem, $\oint_T h(z)dz = 2\pi i c_{-1} = 2\pi i \cdot \frac{1}{64!} = \frac{2\pi i}{64!}$ (we only deal with singularity 0).

C) Coefficient $a_{-1}$ of the term with $z^{-1} = \frac{1}{z} = \frac{1}{z^1}$ in (1) is called the residue of function $f(z)$ at point $z_0 = 0$, here point $z_0 = 0$ is finite.

(notation is the following: $Res_{z=0}(f) = Res_{z=0}(f) = a_{-1}$).

In order to find the coefficient that corresponds to values of $n$ such that $\frac{1}{z^{n^2 - 5n + 1}} = \frac{1}{z^1}$, search for natural solutions to equation

$$n^{2} - 5n = 0 \Rightarrow$$ $$n(n - 5) = 0 \Rightarrow n = 0, n = 5,$$

therefore, coefficient $a_{-1}$ equals

$$\frac{(-1)^0}{(0^2)!} + \frac{(-1)^5}{(5^2)!} = 1 - \frac{1}{25!}.$$

Point $z_0 = 0$ is finite, hence residue $Res_{z=0}f = a_{-1} = 1 - \frac{1}{25!}$

D) Consider

$$\begin{aligned} g(z) &= \sin z \cdot f(z) = \left(z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots + (-1)^n \frac{z^{2n+1}}{(2n+1)!} + \cdots\right) \times \\ &\quad \times \sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{(n^2)!} z^{5n - n^2 - 1} = \left(z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots + (-1)^n \frac{z^{2n+1}}{(2n+1)!} + \cdots\right) \times \\ &\quad \times \sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{(n^2)!} \frac{1}{z^{n^2 - 5n + 1}} \end{aligned}$$

(here we sum up terms $\sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{(n^2)!} \frac{1}{z^{n^2-5n+1}}$, multiplied by terms of

$$z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots + (-1)^n \frac{z^{2n+1}}{(2n+1)!} + \cdots,$$

which lead to the sum of terms with $\frac{z^{2k+1}}{z^{n^2-5n+1}} = \frac{1}{z^{n^2-5n+1-(2k+1)}}$ (according to properties of exponents and power functions), $k$ is an index related to terms of the series

$$z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots, n \text{ is an index related to terms of the series}$$

$\sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{(n^2)!} z^{5n-n^2-1}, k$ is a non-negative integer).

In case of $g(z) = \sin z \cdot f(z)$, $k$ is integer, equate

$$\begin{array}{l} n^2 - 5n - (2k + 1) + 1 = 1 \Rightarrow \\ n^2 - 5n = 2k + 1 \end{array}$$

If $n = 2l$, $l$ is integer, then equation (4) does not have natural solutions, because the left-hand side (i.e. $n^2 - 5n = 4l^2 - 10l$) is even, but the right-hand side (i.e. $2k + 1$) is odd.

If $n = 2l + 1$, $l$ is integer, then

$$\begin{array}{l} (2l + 1)^2 - 5(2l + 1) = 2k + 1 \Rightarrow \\ 4l^2 + 4l + 1 - 10l - 5 = 2k + 1 \Rightarrow \\ 4l^2 + 4l - 10l - 5 = 2k \Rightarrow \\ 4l^2 + 4l - 10l - 2k = 5 \end{array}$$

If $n = 2l + 1$, $l$ is integer, then equation (4) does not have integer solutions either, because the left-hand side (i.e. $4l^2 + 4l - 10l - 2k$) is even, but the right-hand side (i.e. 5) is odd.

It means that on the whole equation (4) does not have integer solutions, therefore, the term of (3) with $z^{-1}$ will not be present, hence the coefficient $d_{-1}$ of the term with $z^{-1} = \frac{1}{z} = \frac{1}{z^1}$ in (4) is zero, which leads to

$$Res_{z=0}(g) = Res_{z=0}(\sin z \cdot f(z)) = d_{-1} = 0$$

(point $z_0 = 0$ is finite).

Coefficient $d_{-1}$ can be computed as $d_{-1} = \frac{1}{2\pi i} \oint_T \frac{g(z)}{(z - 0)^{-1 + 1}} = \frac{1}{2\pi i} \oint_T g(z) dz$, here $T: |z| = (R/2)$.

Thus, by residue theorem, $\oint_T g(z) dz = 2\pi i d_{-1} = 2\pi i \cdot 0 = 0$ (we only deal with singularity 0).

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