# Thread: why does ∫c dz/(1 + z^4) = 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2.

1. ## why does ∫c dz/(1 + z^4) = 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2.

why does ∫c dz/(1 + z^4) = 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2.

2. ## Re: why does ∫c dz/(1 + z^4) = 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2.

What is the countour you are integrating?

3. ## Re: why does ∫c dz/(1 + z^4) = 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2.

C represents

counterclockwise-oriented contour C consisting of the upper half circle of radius R centered at the origin ( γ) and the real axis for x in [-R, R].

4. ## Re: why does ∫c dz/(1 + z^4) = 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2.

Originally Posted by blueyellow
C represents

counterclockwise-oriented contour C consisting of the upper half circle of radius R centered at the origin ( γ) and the real axis for x in [-R, R].
Why don't you use partial fractions decomposition and then just use the residue theorem, or Cauchy's integral formulat, etc.?