A question regarding the order of a pole

[JmsNxn]

Hi, I was wondering about how to determine the residue of a pole that is written in the form:

f(z) = \frac{1}{(1 + t^r)^{\frac{m}{r}}}

Here:
m \in \mathbb{N}\,\,\,\,\;\,\,\,\,r \in \mathbb{R}

And if it's possible, r could be complex.

Would the order of the pole be m? Implying that the r's cancel each other out? Would this give the usual residue theorem:

res(f, e^{\frac{\pi}{r}i}) = \frac{1}{(m-1)!} \lim_{z \to e^{\frac{\pi}{r}i}} \frac{d^{m-1}}{dz^{m-1}} (z - e^{\frac{\pi}{r}i})^{m} f(z)

Or does the residue theorem fail all together here? If so, then, is there any way other way of solving the integral:

\int_{C} f(z)dz

When C is a simple closed contour with the pole inside of it. Any help would be greatly appreciated. I didn't post this in the homework section because it isn't a homework question, just a question of intrigue and research. Thanks.

 

[jackmell]

First start simple. Consider:

f(t)=\frac{1}{(1+\sqrt{t})^2}
and it's integral over a small contour say C=1+1/4 e^{it} looping just once around the pole over an analytic extension of each branch w_k, k=1,2:
\mathop\oint\limits_C f(t)dt=2\pi i r_k
where r_k is the residue for each branch. Can you use that limit expression for those residues keeping in mind that over the branch w_2, the function is actually
f_2(t)=\frac{1}{(1-\sqrt{t})^2}
when \sqrt{t} is interpreted as it's principal value.

 

[JmsNxn]

By the way you seem to phrase it, I'm going to go with 'no'. I guess there requires more work in solving the residue of poles defined in this manner? I'm sorry, but being to new to complex analysis affords me little more reduction of the idea. I still think 'no' though, as hard as I think.

 

[jackmell]

For the case

f_1(t)=\frac{1}{(1+\sqrt{t})^2}
f_2(t)=\frac{1}{(1-\sqrt{t})^2}

the two residues are:

r_1=\lim_{t\to 1}\frac{d}{dt}\left\{(t-1)^2 f_1(t)\right\}=0
r_2=\lim_{t\to 1}\frac{d}{dt}\left\{(t-1)^2 f_2(t)\right\}=2

so that

\mathop\oint\limits_C f_2(t)dt=4\pi i