A harmonic function is analytic if it satisfies the Laplace equation.

Ifu(x,y) = 2x² — 2y² + 4xy is a harmonic function, then its conjugate harmonic function v(x,y) is

The correct option is: 4xy — 2x² + 2y² + constant

Step-by-Step Explanation:

We are given the real part of an analytic function:
$$u(x,y)=2x^2-2y^2+4xy$$
We need to find its harmonic conjugate function $v(x,y)$.

For $u$ and $v$ to be conjugate harmonic functions, they must satisfy the Cauchy-Riemann equations:
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\hspace{1em}\text{and}\hspace{1em}\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

First, let's calculate the partial derivatives of $u(x,y)$ with respect to $x$ and $y$:
$$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(2x^2-2y^2+4xy)=4x+4y$$
$$\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}(2x^2-2y^2+4xy)=-4y+4x$$

Using the first Cauchy-Riemann equation:
$$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=4x+4y$$
Integrate both sides with respect to $y$ keeping $x$ constant:
$$v(x,y)=\int (4x+4y)\partial y=4xy+2y^2+f\left(x\right)$$
where $f\left(x\right)$ is an arbitrary function of $x$.

Now, differentiate our expression for $v(x,y)$ with respect to $x$:
$$\frac{\partial v}{\partial x}=4y+f^{\text{'}}\left(x\right)$$

Using the second Cauchy-Riemann equation ($\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$):
$$4y+f^{\text{'}}\left(x\right)=-(-4y+4x)$$
$$4y+f^{\text{'}}\left(x\right)=4y-4x$$
Subtracting $4y$ from both sides gives:
$$f^{\text{'}}\left(x\right)=-4x$$

Integrating $f^{\text{'}}\left(x\right)$ with respect to $x$ yields:
$$f\left(x\right)=-2x^2+c$$
where $c$ is a constant.

Substituting $f\left(x\right)$ back into the expression for $v(x,y)$:
$$v(x,y)=4xy+2y^2-2x^2+\text{constant}$$
This matches the correct option.