Discussion

Let z be a complex variable. For a counter-clockwise integration around a unit circle C, centred at origin,

\[ \oint_{C} \frac{1}{5z-4}\,dz = A\pi i, \]

the value of A is

Answer: A

The value of A is 2/5.

The integrand has a single pole where 5z - 4 = 0, at z = 4/5, which lies inside the unit circle. Write it as

\[ \frac{1}{5z-4} = \frac{1}{5}\cdot\frac{1}{z - 4/5} \]

so the residue at the pole is 1/5. By the residue theorem,

\[ \begin{aligned} \oint_{C}\frac{dz}{5z-4} &= 2\pi i \times \frac{1}{5} \\ &= \frac{2}{5}\pi i \end{aligned} \]

Hence A = 2/5. Forgetting the factor 1/5 gives A = 2; 4/5 is the location of the pole, not the residue; 1/2 has no route from the integral.

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