Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 5M

Year: Dec 2014

1) Along the path y = x

$\because y = x, dy = dx...............(1)$

We know $z = x + iy$

$\therefore dz = dx + idy$

i.e. $dz = dz + idz = (1 + i)dx$ from (1)

x: 0 to 1

$\therefore \int\limits_{0}^{1 + i} (x^2 + iy)dz = \int\limits_{0}^{1} (x^2 + ix)(1 + i)dx \\ = (1 + i) \bigg[\frac{x^3}{3} + \frac{ix^2}{2}\bigg]_{0}^{1} dx \\ = (1 + i) \bigg(\frac{1}{3} + \frac{i}{2} \bigg) \\ =\frac{(1 + i) ( 2 + 3i)}{6} \\ = \frac{-1 + 5i}{6}$

$$\therefore \int\limits_{0}^{1 + i} (x^2 + iy)dz = \frac{-1 + 5i}{6}$$

2) Along the path $y = x^2$

$\because y = x^2, dy = 2x..................(1)$

We know $z = x + iy$

$\therefore dz = dx + idy$

i.e. $dz = (1 + 2ix)dx$ from (1)

$\therefore \int\limits_{0}^{1 + i} (x^2 + iy)dz = \int\limits_{0}^{1} (x^2 + ix^2)(1 + i2x)dx \\ = \int\limits_{0}^{1} (1 + i) (x^2 + i2x^3)dx \\ = (1 + i) \bigg[\frac{x^3}{3} + 2i\frac{x^4}{4} \bigg]_{0}^{1} \\ = (1 + i) \bigg(\frac{1}{3} + \frac{i}{2} \bigg) \\ = \frac{(1 + i) (2 + 3i)}{6} \\ = \frac{-1 + 5i}{6} $

$$\therefore \int\limits_{0}^{1 + i} (x^2 + iy)dz = \frac{-1 + 5i}{6}$$

The two line integrals are equal.