Question

Consider the following expression: \[ z = \sin(y + it) + \cos(y - it) \] where \( z \), \( y \), and \( t \) are variables, and \( i = \sqrt{-1} \) is a complex number. The partial differential equation derived from the above expression is

• A. \( \frac{\partial^2 z}{\partial t^2} + \frac{\partial^2 z}{\partial y^2} = 0 \)
• B. \( \frac{\partial^2 z}{\partial t^2} - \frac{\partial^2 z}{\partial y^2} = 0 \)
• C. \( \frac{\partial z}{\partial t} - i \frac{\partial z}{\partial y} = 0 \)
• D. \( \frac{\partial z}{\partial t} + i \frac{\partial z}{\partial y} = 0 \)

Hint:

When deriving partial differential equations from functions involving both real and imaginary parts, the key is to differentiate with respect to each variable and apply the appropriate rules.

Answer 1

The Correct Option is A

We are given the expression for \( z \): \[ z = \sin(y + it) + \cos(y - it) \] First, we can differentiate \( z \) with respect to \( t \) and \( y \), then compute the second derivatives.

Step 1: Differentiate with respect to \( t \).
The derivative of \( z \) with respect to \( t \) is: \[ \frac{\partial z}{\partial t} = \cos(y + it) \cdot i + (-\sin(y - it)) \cdot (-i) \] Simplifying this, we get: \[ \frac{\partial z}{\partial t} = i \left[ \cos(y + it) + \sin(y - it) \right] \]

Step 2: Differentiate with respect to \( y \).
The derivative of \( z \) with respect to \( y \) is: \[ \frac{\partial z}{\partial y} = \cos(y + it) - \sin(y - it) \]

Step 3: Compute the second derivatives.
Now, we compute the second derivative of \( z \) with respect to both \( t \) and \( y \): \[ \frac{\partial^2 z}{\partial t^2} = - \left( \sin(y + it) + \cos(y - it) \right) \] and \[ \frac{\partial^2 z}{\partial y^2} = - \left( \sin(y + it) + \cos(y - it) \right) \] Thus, we find: \[ \frac{\partial^2 z}{\partial t^2} + \frac{\partial^2 z}{\partial y^2} = 0 \]

Final Answer: (A) \( \frac{\partial^2 z}{\partial t^2} + \frac{\partial^2 z}{\partial y^2} = 0 \)