NBHM 2020 PART A Question 2 Solution

Let $u$ and $v$ be the real and imaginary parts respectively of the function $f(z) = \frac{1}{z^2 −6z +8}$ of a complex variable $z = x + i y$. Let $C$ be the simple closed curve $|z| = 3$ oriented in the counter clockwise direction. Evaluate the following integral: $$\int_C (udy + vdx).$$
Solution: Let $f(z) = u + iv$ and $dz = dx+idy$, then $$f(z)dz = (u + iv)(dx+idy) = (udx-vdy)+i(udy+vdx).$$ Also, $$\int_C f(z)dz = \int_C ((udx-vdy)+i(udy+vdx))$$ which is equal to $$\big(\int_C udx-vdy\big) + i \big(\int_C udy+vdx\big).$$ So the required integral is the imaginary part of the value of the integral $\int_C \frac{1}{z^2 −6z +8}dz$ which can be calculated using the residue theorem.
Share to your groups:
SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO, SUBSCRIBE TO MY BLOG AND SUGGEST PROBLEMS TO SOLVE.

No comments:

Post a Comment

Featured Post

NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...

Popular Posts