Residue theorem: ∮tan z dz on |z|=2

tan z = sin z / cos z has simple poles where cos z = 0, i.e. z = (2n + 1)π/2.

For C: |z| = 2, the poles inside are z = π/2 ≈ 1.57 and z = −π/2 ≈ −1.57. The next poles at ±3π/2 ≈ ±4.71 are outside.

Residue at a simple pole z = a of p(z)/q(z) is p(a)/q′(a). Here p = sin z and q = cos z, q′ = −sin z, so

Res = sin a / (−sin a) = −1 at every pole.

• At z = π/2: Res = −1
• At z = −π/2: Res = −1

Sum of residues inside C = −2. By Cauchy's residue theorem:

Answer: ∮_C tan z dz = 2πi(−2) = −4πi