We are given \( u – v \). A standard technique involves constructing a new function \( F(z) \) such that its real part corresponds to the given value.

Let us define:

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

Expanding this using \( f(z) = u + iv \):

$$ F(z) = (1+i)(u+iv) = (u – v) + i(u + v) $$

Let \( F(z) = U + iV \). Here, the real part is:

$$ U = u – v = e^x(\cos y – \sin y) $$

Since \( f(z) \) is analytic, \( F(z) \) is also analytic.

We use the Milne-Thomson method to find \( F(z) \) directly from its real part \( U \).

First, find the partial derivatives of \( U \) with respect to \( x \) and \( y \):

1. Derivative with respect to x:

$$ \phi_1(x, y) = \frac{\partial U}{\partial x} = \frac{\partial}{\partial x}[e^x(\cos y – \sin y)] $$ $$ \phi_1(x, y) = e^x(\cos y – \sin y) $$

2. Derivative with respect to y:

$$ \phi_2(x, y) = \frac{\partial U}{\partial y} = e^x \frac{\partial}{\partial y}(\cos y – \sin y) $$ $$ \phi_2(x, y) = e^x(-\sin y – \cos y) = -e^x(\sin y + \cos y) $$

According to the Milne-Thomson method:

$$ F'(z) = \phi_1(z, 0) – i\phi_2(z, 0) $$

Calculate \( \phi_1(z, 0) \):

$$ \phi_1(z, 0) = e^z(\cos 0 – \sin 0) = e^z(1 – 0) = e^z $$

Calculate \( \phi_2(z, 0) \):

$$ \phi_2(z, 0) = -e^z(\sin 0 + \cos 0) = -e^z(0 + 1) = -e^z $$

Substitute these into the expression for \( F'(z) \):

$$ F'(z) = e^z – i(-e^z) $$ $$ F'(z) = e^z + i e^z = e^z(1 + i) $$

Integrate \( F'(z) \) with respect to \( z \):

$$ F(z) = \int e^z(1+i) \, dz $$ $$ F(z) = (1+i)e^z + C $$

Where \( C \) is an arbitrary complex constant.

Now, recall from Step 1 that \( F(z) = (1+i)f(z) \). Substitute this back:

$$ (1+i)f(z) = (1+i)e^z + C $$

Divide the entire equation by \( (1+i) \):

$$ f(z) = e^z + \frac{C}{1+i} $$

Since \( \frac{C}{1+i} \) is just another constant, let’s call it \( c \).