Solving Complex Integral: Cauchy's Formula

[zenite]

1. Integrate z²/(z⁴-1) counterclockwise around x² + 16y²=42. Cauchy's Integral Forumula3. Solution
I found the points z=1,-1,i,-i where the function is not defined. Using partial fractions to split them up, and integral them separately.

Only points z=1,-1 lies in the contour, so...
\oint0.25/(z-1) + 0.25/(z+1) + 1/(z^2+1) dz
= 0.25(2Pi I + 2Pi I) + 0 = Pi I

Ans is 0. can anyone find my mistake?
1. Integrate sinh2z/z⁴ counterclockwise around the unit circle.2. Cauchy's Integral Forumula3. Solution

\oint sinh2z/z^4 = \oint sinh2z/(z-0)^4
= 2*PI*i/3! * (sinh2z)'''

Differentiating sinh2z thrice gives 8cosh2z

Hence, integral at z=0 = (8/3)*PI*i

Ans is (8/3)*PI. Again, can anyone spot my mistake.

 

[latentcorpse]

for the first question:

the residue at z=1 is the limit as z goes to 1 of:

(z-1) \left( 0.25 / (z-1) + 0.25 / (z+1) + 1 / (z^2+1) \right)
put that in and you get 0.25 + (-0.25) = 0

i imagine this will also happen at the z=-1 pole

then just use Cauchy's residue theorem that

\int_\gamma f(z) dz = \displaystyle \sum_i Res(f, c_i) where c_i are the poles of f(z) and you'll get the whole thing to integrate to 0+0=0