Complex Variables and Partial Differential Equations (3130005)

Question-3c ( i )-OR

BE | Semester-3 Winter-2019 | 26-11-2019

Q3) (c ( i ))

(a) |z| = 2

Let,

I = \int_{C} \frac{z + 3}{z - 1} d z

\Rightarrow z_{0} = 1

is non - analytic &

|1| < 2

then

1 \in C

→ By comparing of

\frac{f (z)}{z - z_{0}} & \frac{z + 3}{z - 1}

, We have

f (z) = z + 3

→ By Cauchy's integral formula,

I = 2 πi f (z_{0}) = 2 πi f (i) = 2 πi (1 + 3)

\Rightarrow I = 8 πi

(b) |z| = \frac{1}{2}

Let,

I = \int_{C} \frac{z + 3}{z - 1} d z

\Rightarrow z_{0} = 1

is non - analytic &

|1| > \frac{1}{2}

then

1 \notin C

Then by Cauchy's integral theorem,

I = 0