What is the value of this integral (using the Argument Principle), F(z) is given by $$F(z) = e^zz^{-2}(z-1)(z^2-4)(z+8)^7$$
What is the value of the integral 
$$\int_0^{2\pi} \frac{F'(3e^{i\theta})}{F(3e^{i\theta})}d\theta \space \space ?$$ 
I think the relevant closed curve I am considering is $\gamma = 3e^{i\theta}$.
I want to use the Argument Principle and say that since F(z) has a total of 3 zeroes inside $\gamma$, one each at z=1, z=2, z=-2, and 2 poles inside $\gamma$, both at z=0 , 
then we must have that 3-2 = 1 = $$\frac {1}{2\pi i}\int_0^{2\pi} \frac{F'(3e^{i\theta})}{F(3e^{i\theta})}d\theta \space \space, $$
which implies that  $$\int_0^{2\pi} \frac{F'(3e^{i\theta})}{F(3e^{i\theta})}d\theta \space \space =2\pi i. $$ 
Is this ok, or have I done something foolish and must instead compute this integral directly?  There is an old student solution that I compared my work to, and his work does a direct computation and claims that the Argument Principle cannot be applied because the 1/z term is not analytic in the disk bounded by $\gamma$.  His answer was 15/4 $\pi$.  
Thanks in advance,
 A: You can't both be right, so if the student's direct computation is correct (I got the same value) that must mean your computation is wrong! It's only a matter of figuring out what the student means.
The function $F$ doesn't have to be holomorphic in the interior of the contour - the argument principle applies when the function $F$ is meromorphic, and counts zeros minus poles with multiplicity. It states
$$\frac{1}{2\pi i}\oint \color{Blue}{g(z)}d\log f(z)=\sum g(c),$$
where $d\log f(z)$ is my shorthand for $f'(z)/f(z)\,dz$ and $c$ ranges over all zeros and poles of $f$ in the interior of the contour; each value $g(c)$ is counted according to its multiplicity, where we view poles as zeros with negative multiplicity. (I like to think of this as a sum over "generalized zeros" each with "signed multiplicity.") The issue with applying it to this problem is that
$$\int_0^{2\pi}\frac{F'(3e^{i\theta})}{F(3e^{i\theta})}d\theta=\oint \frac{F'(z)}{F(z)}\frac{dz}{iz}=\oint \color{Blue}{\frac{1}{iz}}d\log F(z)$$
$$=2\pi i\left[-2\frac{1}{i\cdot0}+\frac{1}{i\cdot 1}+\frac{1}{i\cdot2}+\frac{1}{i\cdot(-2)}\right]$$
would involve division by zero! If you naively (and incorrectly) substitute $d\theta=dz$ you will of course obtain the value $2\pi i(-2+1+1+1)$, which seems to be what you did, but $d\theta=dz/(iz)$.
You seem to indicate you have another student's direct computation in front of you to look at; are you able to follow along with the calculation? One should be able to write $(\log F)'$ as a sum of $(z-a)^n$s for some values $a$ and integers $n$, which should be easy to contour integrate.
