How do I evaluate $ \ \frac{1}{2 \pi i} \oint_{|z|=3} \frac{f' (z)}{f(z)}\,dz ,~~\mbox{ where }~ f(z) = \frac{z^2 (z-i )^3 e^z}{3 (z+2 )^4 (3z - 18 )^5} \ $ ?
The singularities are z = -2 and z = 6. But the $ \frac{f' (z)}{f(z)}\ $ part does not make much sense, it looks like L'Hospital rule but it's really not. It'd be too tedious if I just calculate $ \frac{f' (z)}{f(z)}\ $. In fact I'm not even sure if I can... It also looks like I might want to use residue theorem, but why $ \frac{f' (z)}{f(z)}\ $?
I am confused about both why this question is asked this way and how to solve it. In the mean time it'll be great if somebody can just tell me how to solve it...
Thanks.
Edit: It might be way way easier to just use Argument principle by Cauchy. But I'm not sure about how to count Poles and Zeros. I mean, poles I know, but WHAT ARE ZEROS? Could somebody solve the problem using this theorem? Thanks!