Show that C (z - z 1 ) -1 (z - z 2 ) -1 dz

Question:

Show that ∫C (z - z1)-1 (z - z2)-1dz = 0 for a simple closed path C enclosing z1 and z2, which are arbitrary.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: