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
Probability and complex function
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