please help: Find all the poles and multiplicities of..

**linalg123** · August 29th 2012, 10:10 PM

Find all the poles and multiplicities of

a) $\left(\frac{4z+6}{z^2+6z+9}\right)^3$

b) $\frac{6z}{z^2+5z+6}+\frac{1}{z+2}$

Attempt:

a) ok so $\left(\frac{4z+6}{z^2+6z+9}\right)^3$

$= \left(\frac{2z+3}{(z+3)^2}\right)^3$

Would that mean pole at z= -3 of order 2 or order 6?

b) $\frac{6z}{z^2+5z+6}+\frac{1}{z+2}$

$= \frac{6z}{(z+2)(z+3)}+ \frac{1}{z+2}$

So there is a pole at z=-3 of order 1 but is the pole at z=-2 of order 1 or order 2?

**Vlasev** · August 29th 2012, 10:31 PM

For part a) this should help

$\frac{1}{(z-1)^4}$

has pole of order 4. What's the order of the pole of this

$\left(\frac{1}{(z-1)^2}\right)^2$ ?

For part b) you are not done. What's the definition of pole? Does your function look like the definition in its current form? I mean, what you have is right but it's more technically correct to write it all out as a fraction of polynomials.

**linalg123** · August 29th 2012, 11:02 PM

thats what i was asking :P
We have covered the first example you showed, but not where the whole fraction is raised to a power. My feeling is that it would also be 4 because they are equivalent.

oh so for part b, 7z+3/ (z+2)(z+3) and then pole at z= -2,-3 both of order 1?

Thanks for the speedy reply

**Vlasev** · August 31st 2012, 10:42 AM

Yes, for part b it's like that.

Your feeling is right. The pole multiplicities are the same. Do you know the limit definition of a pole of order $n$ ?

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