Given:

$v=e^x.\sin y$

Since we have to prove that v is harmonic function.

The function said to be Harmonic,

If

  $\dfrac{\mathrm{∂}^2 u}{\mathrm{∂} x^2}+\dfrac{\mathrm{∂}^2 u }{\mathrm{∂} y^2}=0$

or

  $\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} x^2}+\dfrac{\mathrm{∂}^2 v }{\mathrm{∂} y^2}=0$

Here v is given.

Therefore we have to show that 

 $\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} x^2}+\dfrac{\mathrm{∂}^2 v }{\mathrm{∂} y^2}=0$

Taking partial derivative of v with respect to y.

$\dfrac{\mathrm{∂} v}{\mathrm{∂} y} =\dfrac{\mathrm{∂} }{\mathrm{∂} y} e^x.\sin y$

$\dfrac{\mathrm{∂} v}{\mathrm{∂} y}= e^x.\cos y $______________$\Psi_1(x,y)$

Again taking partial derivative with respect to y.

$\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} y^2} =\dfrac{\mathrm{∂}^2 }{\mathrm{∂} y^2} e^x.\cos y$

$\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} y^2} =e^x.(-\sin y)$

$\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} y^2} =-e^x.\sin y \cdots\cdots(1)$

Taking partial derivative fo v with respect to x.

$\dfrac{\mathrm{∂} v}{\mathrm{∂} x} =\dfrac{\mathrm{∂} }{\mathrm{∂} x} e^x.\sin y$

$\dfrac{\mathrm{∂} v}{\mathrm{∂} x} =e^x.\sin y$______________$\Psi_2(x,y)$

Again taking partial derivative with respect to x.

$\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} y^2} =\dfrac{\mathrm{∂}^2 }{\mathrm{∂} y^2} e^x.\sin y$

$\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} y^2} =e^x.\sin y$_____________(2)

Add 1 & 2

$\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} x^2}+\dfrac{\mathrm{∂}^2 v }{\mathrm{∂} y^2}=e^x.\sin y+(-e^x.\sin y)$

$\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} x^2}+\dfrac{\mathrm{∂}^2 v }{\mathrm{∂} y^2}=e^x.\sin y-e^x.\sin y$

 $\dfrac{\mathrm{∂}^2 v}{\mathrm{∂} x^2}+\dfrac{\mathrm{∂}^2 v }{\mathrm{∂} y^2}=0$

Hence v is Harmonic Function.

Now we have to find its corresponding harmonic conjugate i.e. "u" & analytic function "f(z)".

So first we will find f(z) using Milne Thomson Method

Using Milne-Thomson-Method

$f'(z) = Ψ_1(z,0) + i\Psi_2(z,0)$

$=e^z.\cos (0)+i.e^z.\sin (0)$ 

$ = e^z(1)+i(0) $ _________________$\because \sin 0=0 , \cos 0=1$

$f'(z)= e^z$

Integrating both sides with respect to z

$∫ f'(z).dz = ∫ e^z.dz$

$f(z) = e^z+c$______This is required analytic function.

We know that   $f(z)=w=u+iv$  &  $z=x+iy$.

Putting these values in above analytic function we get,

$u+iv= e^{(x+iy)}+c$

           $ = e^x.e^{iy} +c$                                                         $\because e^{a+b}=e^a.e^b$

           $ = e^x.(\cos y + i\sin y) +c$                          $ \because e^{iθ}= \cos y +i\sin y$

$u+iv= e^x\cos y + ie^x\sin y+c$

Comparing R.H.S with L.H.S

we get imaginary part v as 

 $v=e^x\sin y$__________Given in Question.

& real part u as

    $u= e^x\cos y +c$

Hence required harmonic conjugate is $\underline {u = e^x\cos y+c}$, & analytic function is $\underline{f(z) = e^z+c} $.