What Value of b Makes the Complex Function Satisfy the 3D Wave Equation?

[Saraphim]

Homework Statement

Given a complex function

$$\tilde{f}(\mathbf{r},t)=Aexp(i(b(x+y)-wt+d))$$

I am asked to determine for what value of b this satisfies the three-dimensional wave equation. x and y are components of r.

Homework Equations

3d wave equation:
$$\nabla^2 f=\frac{1}{v^2} \frac{d^2 f}{dt^2}$$

The Attempt at a Solution

I take $$f=Re(\tilde{f})=Acos(b(x+y)-wt+d)$$ then write up the three-dimensional wave equation

$$\nabla^2 f=\frac{1}{v^2} \frac{d^2 f}{dt^2}$$
$$-2Ab^2\cos(b(x+y)-wt+d)=-\frac{1}{v^2}w^2A\cos(b(x+y)-wt+d)$$

Now, it is simply a matter of isolating b in the final equation. Is this correct?

I get $$b=\frac{w}{v \sqrt{2}}$$

If this is correct, could I have reached this conclusion in an easier manner? I have this nagging feeling I'm missing out on all kind of complex goodness by just going to real parts immediately.

EDIT EDIT: Total nonsense answer before, sorry. Dunno how I missed that minus.

 

[anathema]

it is important to carefully check your work and make sure that it is correct before presenting it to others. In this case, your solution is incorrect. I will guide you through the correct process to solve this problem.

First, let's start with the three-dimensional wave equation:

\nabla^2 f=\frac{1}{v^2} \frac{d^2 f}{dt^2}

We can rewrite this equation as:

\nabla^2 f - \frac{1}{v^2} \frac{d^2 f}{dt^2} = 0

Now, let's substitute in the given complex function:

\nabla^2 (Ae^{i(b(x+y)-wt+d)}) - \frac{1}{v^2} \frac{d^2 (Ae^{i(b(x+y)-wt+d)})}{dt^2} = 0

Using the chain rule, we can simplify the second term:

\nabla^2 (Ae^{i(b(x+y)-wt+d)}) + \frac{1}{v^2} w^2 (Ae^{i(b(x+y)-wt+d)}) = 0

Now, let's expand the Laplacian operator:

(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2})(Ae^{i(b(x+y)-wt+d)}) + \frac{1}{v^2} w^2 (Ae^{i(b(x+y)-wt+d)}) = 0

We can now simplify this equation by using the fact that the exponential term is independent of x, y, and z:

-A(b^2+w^2)\cos(b(x+y)-wt+d) + \frac{1}{v^2} w^2 A \cos(b(x+y)-wt+d) = 0

Simplifying further, we get:

-A(b^2+w^2+\frac{w^2}{v^2})\cos(b(x+y)-wt+d) = 0

Since the cosine term cannot be equal to zero (unless A=0), we must have:

b^2+w^2+\frac{w^2}{v^2} = 0

Solving for b, we get:

b = \pm \