Question

Let $z$ be a complex variable. For a counter-clockwise integration around a unit circle $C$. centred at origin.
$I={\oint }_c\frac{1}{5z-4}dz=A\pi i$
the value of $A$ is

• A. $2/5$
• B. $1/2$
• C. $2$
• D. $4/5$

Answer 1

The correct option is A $2/5$

$C:|z|=1$ and $f\left(z\right)=\frac{1}{5z-4}$
Pole of $f\left(z\right)$ is $z=\frac{4}{5}$,a simple pole
Pole $z=\frac{4}{5}$ lies inside the circle

$Resf(z{)}_{z=4/5}={lim}_{z\to 4/5}(z-\frac{4}{5})f(z)$
$={lim}_{z\to 4/5}\left(\frac{1}{5}\right)(5z-4)\frac{1}{(5z-4)}$
$={lim}_{z\to 4/5}\left(\frac{1}{5}\right)=\frac{1}{5}$
B Cauchy's Residue theorem
$I={\oint }_{C}f\left(z\right)dz$
$=2\pi i$ (sum of residues)
$=2\pi i\left[\frac{1}{5}\right]$
${\oint }_C\left(\frac{1}{5z-4}\right)dz=\frac{2\pi i}{5}=A\pi i$ (given)
Hence on comparison $A=\frac{2}{5}=0.4$