Yes, the function $g(z) = v(x, y) + i u(x, y)$ can also be an analytic function, provided that $u(x, y)$ and $v(x, y)$ satisfy the Cauchy-Riemann equations.

Explanation:

Let $f(z) = u(x, y) + i v(x, y)$ be an analytic function, where $z = x + iy$, and $u(x, y)$ and $v(x, y)$ are the real and imaginary parts of $f(z)$, respectively.

For $f(z)$ to be analytic, the Cauchy-Riemann equations must hold:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

Now consider the function $g(z) = v(x, y) + i u(x, y)$. For $g(z)$ to be analytic, the real part $v(x, y)$ and the imaginary part $u(x, y)$ must satisfy the Cauchy-Riemann equations in the same form, but with the roles of $u$ and $v$ swapped:

$\frac{\partial v}{\partial x} = \frac{\partial u}{\partial y}, \quad \frac{\partial v}{\partial y} = -\frac{\partial u}{\partial x}$

If $u(x, y)$ and $v(x, y)$ satisfy the original Cauchy-Riemann equations for $f(z)$, then these new equations for $g(z)$ are also satisfied because:

1. $\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$
2. $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}$

Thus, $g(z)$ can indeed be an analytic function.

Would you like a detailed derivation of the Cauchy-Riemann equations or further clarification on any part of this?

Here are some related questions:

1. What are the necessary conditions for a function to be analytic?
2. Can you provide examples of functions that are analytic everywhere?
3. How do the Cauchy-Riemann equations relate to harmonic functions?
4. What is the significance of the Cauchy-Riemann equations in complex analysis?
5. How would the result change if $u(x, y)$ and $v(x, y)$ were swapped initially?

Tip: Remember that the Cauchy-Riemann equations are not only necessary but also sufficient conditions for differentiability in complex analysis, provided the partial derivatives are continuous.