Question

The value of k that makes the complex-valued function \[ f(z) = e^{-kx} (\cos 2y - i \sin 2y) \] analytic, where \(z = x + iy\), is ................ (Answer in integer)

Hint:

When a complex function involves exponential and trigonometric terms, try to combine them using Euler's formula (\(e^{i\theta} = \cos\theta + i\sin\theta\)). If you can manipulate the resulting expression into a function of \(z = x+iy\) alone, then it's analytic, and you can find the required parameters. This is often faster than using the C-R equations.

Answer 1

Step 1: Understanding the Concept:
A complex function \(f(z) = u(x, y) + iv(x, y)\) is analytic in a region if it is differentiable at every point in that region. A necessary condition for a function to be analytic is that its real part \(u\) and imaginary part \(v\) must satisfy the Cauchy-Riemann (C-R) equations.
Step 2: Key Formula or Approach:
1. Express the function \(f(z)\) in the form \(u(x, y) + iv(x, y)\).
2. The Cauchy-Riemann equations are: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{and} \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] 3. Calculate the partial derivatives and apply the C-R equations to find the value of \(k\).
Step 3: Detailed Calculation:
1. Identify u and v:
The given function is \(f(z) = e^{-kx} \cos 2y - i e^{-kx} \sin 2y\).
Comparing this to \(f(z) = u + iv\), we have: \[ u(x, y) = e^{-kx} \cos 2y \] \[ v(x, y) = -e^{-kx} \sin 2y \] 2. Calculate the partial derivatives:
- \(\frac{\partial u}{\partial x} = -k e^{-kx} \cos 2y\)
- \(\frac{\partial u}{\partial y} = e^{-kx} (-\sin 2y) \cdot 2 = -2 e^{-kx} \sin 2y\)
- \(\frac{\partial v}{\partial x} = -(-k e^{-kx}) \sin 2y = k e^{-kx} \sin 2y\)
- \(\frac{\partial v}{\partial y} = -e^{-kx} (\cos 2y) \cdot 2 = -2 e^{-kx} \cos 2y\)
3. Apply the first C-R equation (\(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\)): \[ -k e^{-kx} \cos 2y = -2 e^{-kx} \cos 2y \] For this to be true for all \(x\) and \(y\), the coefficients must be equal: \[ -k = -2 \implies k = 2 \] 4. Apply the second C-R equation (\(\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\)) to verify: \[ -2 e^{-kx} \sin 2y = -(k e^{-kx} \sin 2y) \] \[ -2 e^{-kx} \sin 2y = -k e^{-kx} \sin 2y \] Again, for this to be true for all \(x\) and \(y\), we must have: \[ -2 = -k \implies k = 2 \] Both equations yield the same result.
Alternative Method using Euler's Formula:
The function can be written as: \[ f(z) = e^{-kx} (\cos 2y - i \sin 2y) = e^{-kx} e^{-i2y} = e^{-kx - i2y} = e^{-(kx + i2y)} \] For this function to be an analytic function of \(z = x+iy\), it must be expressible as a function of \(z\) alone. Let's try to relate the exponent to \(z\): \[ kx + i2y \] If we set \(k=2\), the exponent becomes \(2x + i2y = 2(x+iy) = 2z\).
Then the function becomes: \[ f(z) = e^{-2z} \] The function \(e^{-2z}\) is an entire function (analytic everywhere), so \(k=2\) is the correct value.
Step 4: Final Answer:
The value of k is 2.
Step 5: Why This is Correct:
Both the application of the Cauchy-Riemann equations and the direct manipulation of the exponential form show that \(k\) must be equal to 2 for the function to be analytic.