Probability and complex function: Unit III: Analytic functions

Example Solved Problems based on harmonic conjugate

Analytic functions

Probability and complex function: Unit III: Analytic functions : Problems based on harmonic conjugate

PROBLEMS BASED ON HARMONIC CONJUGATE

 

Example 3.2.1. If f (z) = ez then show that u and v are harmonic functions.

Solution: Given: f(z) = ez

u + iv = ex + iy = ex eiy

 = ex [cos y + i sin y]

= ex cos y + iex sin y


uxx + uyy = ex cos y – ex cos y

= 0

vxx + vyy = ex siny  - ex sin y

= 0

Second order partial derivatives of u and v are continuous.

So, u and v are harmonic function.

 

Example 3.2.2. Give an example such that u and v are harmonic but u + iv is not analytic. [A.U. N/D 2005]

Solution :

Note : u = Re z2

u = x2 - y2, v = -y/x2 + y2  v = im 1/z

 

Example 3.2.3 Find the value of m if u = 2x2- my2+ 3x is harmonic. [A.U N/D 2016 R-13]

Solution: Given: u = 2x2 - my2 + 3x

2u/∂x2 + ∂2u/∂y2 = 0  [ u is harmonic] .. (1)


(1) (4) + (-2m) = 0

⇒ m = 2

 

Probability and complex function: Unit III: Analytic functions : Tag: : Analytic functions - Example Solved Problems based on harmonic conjugate