DrMath said:
So z=0 is actually a suspected pole just by looking at the question.
Yes the denominator of your fraction is zero there, so you need to test for a pole at [itex]z=0[/itex].
DrMath said:
However - it is not a real pole right?
Because my test of pole procedure determines it is not a pole.
No it
is a real pole. As I said in my previous post, if the limit of the function multiplied by [itex]z[/itex] as [itex]z\rightarrow 0[/itex] was zero then it would be removable.
You have multiplied by [itex]z^{15}[/itex].
That's not the same thing!
DrMath said:
From my understanding from you; it is a removable singularity if we take the limit of
z tends to 0 (pole) --- from my attachment i raise it to the power of 15. Is this not correct ?
No it's not at all correct.
The
order of a pole at some point [itex]z_0[/itex] is defined as the
smallest integer [itex]m[/itex] you can find such that:
[tex]
\lim_{z\rightarrow z_0}{\left[\left(z-z_0\right)^{m+1}\,f\!\left(z\right)\right]=0[/tex]
If [itex]m[/itex] is zero then it is called a
removable singularity. (If [itex]m[/itex] is less than zero then it's not a pole at all)
Given this, can you now see what the order of your pole is?
PS My initial post contained a mistake in my description of the limit you are supposed to take, which probably hasn't helped with your confusion! I'll edit it and fix it...