Show that 𝑣(𝑥, 𝑦) = 𝑒 𝑥 sin(𝑦) is a harmonic function. Find 𝑢(𝑥, 𝑦) such that 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic.

To show that \( v(x, y) = e^x \sin(y) \) is a harmonic function and to find \( u(x, y) \) such that \( f(z) = u + iv \) is analytic, we follow these steps:

### Step 1: Verify that \( v(x, y) = e^x \sin(y) \) is a harmonic function

A function \( v(x, y) \) is harmonic if it satisfies the Laplace equation:
\[
\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0
\]

Let's compute the second partial derivatives of \( v(x, y) = e^x \sin(y) \):

1. Compute \( \frac{\partial v}{\partial x} \):
   \[
   \frac{\partial v}{\partial x} = \frac{\partial}{\partial x}(e^x \sin(y)) = e^x \sin(y)
   \]

2. Compute \( \frac{\partial^2 v}{\partial x^2} \):
   \[
   \frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x}(e^x \sin(y)) = e^x \sin(y)
   \]

3. Compute \( \frac{\partial v}{\partial y} \):
   \[
   \frac{\partial v}{\partial y} = \frac{\partial}{\partial y}(e^x \sin(y)) = e^x \cos(y)
   \]

4. Compute \( \frac{\partial^2 v}{\partial y^2} \):
   \[
   \frac{\partial^2 v}{\partial y^2} = \frac{\partial}{\partial y}(e^x \cos(y)) = -e^x \sin(y)
   \]

Now, add these second derivatives:
\[
\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = e^x \sin(y) - e^x \sin(y) = 0
\]

Since the sum of the second partial derivatives is zero, \( v(x, y) = e^x \sin(y) \) is indeed a harmonic function.

### Step 2: Find \( u(x, y) \) such that \( f(z) = u + iv \) is analytic

For \( f(z) = u(x, y) + iv(x, y) \) to be analytic, the functions \( u(x, y) \) and \( v(x, y) \) must satisfy the Cauchy-Riemann equations:
\[
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
\]

Given \( v(x, y) = e^x \sin(y) \), we have already computed:
\[
\frac{\partial v}{\partial y} = e^x \cos(y) \quad \text{and} \quad \frac{\partial v}{\partial x} = e^x \sin(y)
\]

According to the Cauchy-Riemann equations, we need:
\[
\frac{\partial u}{\partial x} = e^x \cos(y)
\]
\[
\frac{\partial u}{\partial y} = -e^x \sin(y)
\]

Now, integrate \( \frac{\partial u}{\partial x} \) with respect to \( x \) to find \( u(x, y) \):
\[
u(x, y) = \int e^x \cos(y) \, dx = e^x \cos(y) + \phi(y)
\]
where \( \phi(y) \) is an arbitrary function of \( y \).

Next, use \( \frac{\partial u}{\partial y} = -e^x \sin(y) \) to determine \( \phi(y) \):
\[
\frac{\partial u}{\partial y} = -e^x \sin(y) + \phi'(y) = -e^x \sin(y)
\]

This implies:
\[
\phi'(y) = 0 \quad \Rightarrow \quad \phi(y) = C
\]
where \( C \) is a constant.

Thus, \( u(x, y) = e^x \cos(y) + C \). We typically take \( C = 0 \) for simplicity, giving:
\[
u(x, y) = e^x \cos(y)
\]

### Conclusion

The analytic function \( f(z) \) corresponding to the given harmonic function \( v(x, y) = e^x \sin(y) \) is:
\[
f(z) = u(x, y) + iv(x, y) = e^x \cos(y) + i e^x \sin(y) = e^x (\cos(y) + i\sin(y))
\]
This simplifies to:
\[
f(z) = e^{x + iy} = e^z
\]
where \( z = x + iy \).

Would you like more details on any step or further questions?

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### Related Questions:
1. What are the Cauchy-Riemann equations, and why are they important?
2. How can we determine if a complex function is harmonic?
3. What are the physical interpretations of harmonic functions?
4. Can we find the harmonic conjugate for other functions?
5. How does the complex exponential function \( e^z \) behave in the complex plane?

### Tip:
When dealing with harmonic functions, remember that their real and imaginary parts are often interconnected through the Cauchy-Riemann equations. This relationship ensures that the overall function is analytic.

Learn how to show that 𝑣(𝑥, 𝑦) = 𝑒 𝑥 sin(𝑦) is a harmonic function and find 𝑢(𝑥, 𝑦) such that 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic. This problem involves advanced concepts in harmonic functions, analyticity, and the application of the Cauchy-Riemann equations. Suitable for advanced undergraduate students and higher levels of study.

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