# complex nos multiplicity

• Nov 22nd 2006, 03:55 AM
edgar davids
complex nos multiplicity
can someone help me with this question pls think i undersdtand trhe concepts but cant quite get the answer thanks

Edgar

Determine the multiplicity of the roots of the following functions:
(a) (z^4 − 1) sin(pi.z)
(b) sin^3(1/z), z does not equal 0
(c) 1 + exp(2z)
• Nov 22nd 2006, 08:28 AM
CaptainBlack
Quote:

Originally Posted by edgar davids
can someone help me with this question pls think i undersdtand trhe concepts but cant quite get the answer thanks

Edgar

Determine the multiplicity of the roots of the following functions:
(a) (z^4 − 1) sin(pi.z)

$
(z^4-1) \sin(\pi\ z)=0
$

has roots when $z^4-1=0$ and/or $\sin(\pi\ z)=0$

The roots of $z^4-1=0$ are $1,\ -1,\ i,\ -i$ all of multiplicity 1

The roots of $\sin(\pi z)$ are $0,\ \pm1,\ \pm2,\ ...$ again each of multiplicity 1.

The roots $1,\ -1$ occur in both terms so they are roots of multiplicity 2 for the original function, all the other roots are roots of multiplicity 1 of the original function.

RonL
• Nov 22nd 2006, 09:09 AM
edgar davids
thanks and a question
firstly thanks captain black for that

second: can yuo pls help me with parts b and c and if poss =give me somehints as to the method

Edgar
• Nov 22nd 2006, 09:41 AM
ThePerfectHacker
Quote:

Originally Posted by edgar davids
(c) 1 + exp(2z)

You want,
$1+e^{2z}=0$
Thus,
$e^{2z}=-1$
By Euler's formula,
$e^{\pi i}=0$
Thus,
$e^{\pi i+2\pi k}=0$
Thus,
$z=2\pi k+\pi i, k\in \mathbb{Z}$
No two are equal thus, each has multiplicity one 1
• Nov 22nd 2006, 09:57 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
You want,
$1+e^{2z}=0$
Thus,
$e^{2z}=-1$
By Euler's formula,
$e^{\pi i}=0$

Typo should be -1

Quote:

Thus,
$e^{\pi i+2\pi k}=0$
Thus,
$z=2\pi k+\pi i, k\in \mathbb{Z}$
No two are equal thus, each has multiplicity one 1
RonL
• Nov 22nd 2006, 10:05 AM
CaptainBlack
Quote:

Originally Posted by edgar davids
can someone help me with this question pls think i undersdtand trhe concepts but cant quite get the answer thanks

Edgar

Determine the multiplicity of the roots of the following functions:
(a) (z^4 − 1) sin(pi.z)
(b) sin^3(1/z), z does not equal 0
(c) 1 + exp(2z)

A general if laborious method to determine the multiplicity of a root is as follows:

Let $z_0$ be a root of a function $f(z)$ analytic in some neighbourhood of $z_0$. Expand $f(z)$ as a Taylor series about $z_0$ then if $k$ is the smallest exponent of $z-z_0$ for which the coefficient is non-zero then $k$ is the multiplicity of the root.

This of course may be recast as the multiplicity of the root is the order of the lowest order derivative of the function at $z_0$ which is non-zero.

Applying this method to (b) shows that each of the roots $z=1/n \pi,\ n=\pm 1,\ \pm 2,\ ...$ has multiplicity 3 as the first and second derivatives of $\sin^3(1/z)$ are zero at these points, but the third derivative is non-zero at these points.

(This method may also be applied to (c), which will confirm ImPerfectHacker conclusion)

RonL
• Nov 22nd 2006, 10:06 AM
ThePerfectHacker
The second problem is $z\not = 0$
We have,
$\sin^3 (1/z)=0$
Using the field property of zero,
If and only if,
$\sin (1/z)=0$

We know that,
$\sin (1/z)=\frac{e^{i/z}-e^{-i/z}}{2i}$
We require that,
$e^{i/z}-e^{-i/z}=0$
Thus,
$e^{i/z}=e^{-i/z}$
Since the complex exponential is no-where zero we can divde through by it obtaining a biconditional statement,
$e^{2i/z}=1$
By Euler's Formula,
$e^{2\pi i}=1$
Thus,
$2i/z=2\pi i+2\pi k$
Thus,
$1/z=-\pi+\pi i k$
Thus,
$z=\frac{1}{\pi i k-\pi }$
It seems all solutions are of multiplicity of order 1.