Question

If u=cosxcoshy, then the analytic function f(z) =u+iv is​

Answer 1

f(z)=u+IV is called penicillium

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$u = cosx \: coshy$

TO DETERMINE

The harmonic conjugate v of u and the analytic function f(z) = u + iv

CALCULATION

$u = cosx \: coshy$

Now

$\displaystyle \: \frac{\partial u}{\partial x} = - sinx \: coshy$

$\displaystyle \: \frac{\partial u}{\partial y} = cosx \: sinh \: y$

$\displaystyle \: \frac{\partial^2 u}{\partial x^2} = - cosx \: cosh \: y$

$\displaystyle \: \frac{\partial^2 u}{\partial y^2} = cosx \: cosh \: y$

So

$\displaystyle \: \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$

So u satisfies Laplace Equation

Hence u is a harmonic function

Let v be the harmonic conjugate of u

Then v satisfies Cauchy - Riemann partial differential equations :

$\displaystyle \: \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \: \: \: and \: \: \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \: \: .....(1)$

Now

$dv = \displaystyle \: \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy$

$\implies \: dv = \displaystyle \: - \frac{\partial u}{\partial y} \: dx + \frac{\partial u}{\partial x} \: dy$

$\implies \: dv = \displaystyle \: - cosx \: sinhy \: \: dx - sinx \: cosh \: y \: \: \: dy$

On Integration we get

$v = - sinx \: cosh \: y \: + c$

Where c is constant

Which is the harmonic conjugate of u

Now

$f(z) = u \: + iv$

$\implies \: f(z) = \displaystyle \: cosx \: cosh \: y \: \: + i( - sinx \: cosh \: y \: + c)$

$\implies \: f(z) = \displaystyle \: cosx \: cosh \: y \: \: - i \: sinx \: cosh \: y \: + ic$

$\implies \: f(z) = \displaystyle \: cos(x + iy) + ic$

$\implies \: f(z) = \displaystyle \: cosz + ic$

Which is the required analytic function

RESULT

1. The harmonic conjugate v of u is

$- sinx \: cosh \: y \: + c$

2.The analytic function

$f(z) = \displaystyle \: cosz + ic$