If f(z) is analytic in a simply connected domain D, then for every closed path C and D

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  1. \(\mathop \oint \limits_C f\left( z \right)dz = 1\)
  2. \(\mathop \oint \limits_C f\left( z \right)dz = 0\)
  3. \(\mathop \oint \limits_C f\left( z \right)dz \ne 0\)
  4. \(\mathop \oint \limits_C f\left( z \right)dz \ne 1\)

Answer (Detailed Solution Below)

Option 2 : \(\mathop \oint \limits_C f\left( z \right)dz = 0\)

Detailed Solution

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Explanation:

Cauchy's Theorem:

If f(z) is single-valued and an analytic function of z and f'(z) is continuous at each point within and on the closed curve c, then according to the theorem, \(\mathop \oint \limits_C f\left( z \right)dz = 0\).

Cauchy's Integral Formula:

For Simple Pole:

If f(z) is analytic within and on a closed curve c and if a (simple pole) is any point within c, then

\(\oint_{c}^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)\)

For Multiple Pole:

If f(z) is analytic within and on a closed curve c, and if a (multiple poles) are points within c, then

\(\oint_{c}^{}\frac{f(z)}{(z-a)^n}dz=\frac{2\pi i}{(n-1)!}\left ( \frac{d^{n-1}f(z)}{dz^{n-1}} \right )_{z\;=\;a}\)

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