Question

u+v=sin2x/cosh2y-cos2x find the analytic function f(z) =u+iv​

Answer 1

Given:

u + v = sin2x / (cosh2y-cos2x)

f(z) = u + iv​

To find:

Find the analytic function f(z) = u + iv​

Solution:

f(z) = u + iv​ be any analytic function

∴if(z) = iu + i²v

∵i² = -1

if(z) = iu - v

Now, f(z) + if(z) = u + iv​ + iu - v

(1 + i)f(z) = (u -v) + i(u + v)

Let, F(z) = U + iV

where F(z) = (1 + i)f(z)

U = u - v

V = u + v

Since, V = u + v = sin2x / (cosh2y-cos2x)

Now,

$\frac{\delta V}{\delta x} = \frac{(cosh2y - cos2x).2cos2x-sin2x(2sinh2y+2sin2x)}{(cosh2y - cos2x)^{2} }$

So, Ψ₂(z , 0) = [δV / δx]₍z , ₀₎

$\psi_2(z,0)= \frac{(cosh0 - cos2z).2cos2z-sin2z(2sinh0+2sin2z)}{(cosh0 - cos2z)^{2} }\\\\\psi_2(z,0) = \frac{(1-cos2z).2cos2-2sin^{2}2z}{(1-cos2z)^{2}} \\\\\psi_2(z,0) = \frac{2cos2-2cos^{2}2z-2sin^{2}2z}{(1-cos2z)^{2}}\\\\\psi_2(z,0) = \frac{2cos2-2(cos^{2}2z+sin^{2}2z)}{(1-cos2z)^{2}}\\\\\psi_2(z,0) = \frac{2cos2-2}{(1-cos2z)^{2}}\\\\\psi_2(z,0) = \frac{-2(1-cos2z)}{(1-cos2z)^{2}}\\\\\psi_2(z,0) = \frac{-2}{1-cos2z}\\\\\psi_2(z,0) = \frac{-2}{1- 1 + 2sin^{2}z}$

$\psi_2(z,0) = \frac{-2}{2sin^{2}z}$

Ψ₂(z , 0) = -cosec²z

And,

$\frac{\delta V}{\delta y} = \frac{(cosh2y - cos2x).0-2sin2x(2sinh2y+0)}{(cosh2y - cos2x)^{2} }$

Ψ₁(z , 0) = [δV / δy]₍z , ₀₎

Ψ₁(z , 0) = 0

By Milne's Method

$F(z) = \int\limits {[0 - icosec^{2}z]} \, dz + c\\\\F(z) = -i\int\limits {cosec^{2}z} \, dz + c$

F(z) = i cotz + c

∵ F(z) = (1 + i)f(z)

(1 + i)f(z) = i cotz + c

$f(z) = \frac{i}{1+i}cotz + \frac{c}{1+i}$

Let c/(1+i) = C'

f(z) = (1/1+i)cotz + C'

$f(z) = \frac{i(1+i)cotz}{(1+i)(1-i)} + C'\\\\f(z) = (\frac{i - i^{2}}{1 - i^{2}}) cotz + C'\\\\f(z) = \frac{i+1}{2}cotz + C'$

f(z) = (i +1 / 2)cotz + C'