Question

Which of the following function satisfies Cauchy-Riemann equations:

• A. $x - iy$
• B. $e^x \cdot e^{-iy}$
• C. $e^y \cdot e^{ix}$
• D. $e^x \cdot e^{iy}$

Hint:

Note that $e^x \cdot e^{iy} = e^{x+iy} = e^z$, which is the standard complex exponential function $f(z) = e^z$. Any function depending purely on $z$ (and not $\bar{z}$) satisfies C-R equations!

Answer 1

The Correct Option is D

Step 1: Concept
For a complex function $f(z) = u(x, y) + i v(x, y)$, the Cauchy-Riemann (C-R) equations are the necessary conditions for $f(z)$ to be complex-differentiable: \[ u_x = v_y \quad \text{and} \quad u_y = -v_x \]

Step 2: Key Formulas and Approach
Express each candidate function as $u(x, y) + i v(x, y)$, calculate partial derivatives $u_x, u_y, v_x, v_y$, and check if both C-R equations hold simultaneously.

Step 3: Step-by-step Explanation

• Testing Option (A): $f(z) = x - iy$
$u = x, v = -y \implies u_x = 1, v_y = -1$. Since $u_x \neq v_y$ ($1 \neq -1$), C-R equations fail.

• Testing Option (B): $f(z) = e^x e^{-iy = e^x \cos y - i e^x \sin y$}
$u = e^x \cos y, v = -e^x \sin y$. $u_x = e^x \cos y$ and $v_y = -e^x \cos y$. Since $u_x \neq v_y$, C-R equations fail.

• Testing Option (C): $f(z) = e^y e^{ix = e^y \cos x + i e^y \sin x$}
$u = e^y \cos x, v = e^y \sin x$. $u_x = -e^y \sin x$ and $v_y = e^y \sin x$. Since $u_x \neq v_y$, C-R equations fail.

• Testing Option (D): $f(z) = e^x e^{iy = e^{x+iy} = e^z = e^x \cos y + i e^x \sin y$}
$u(x, y) = e^x \cos y$ and $v(x, y) = e^x \sin y$. Compute partial derivatives: \[ u_x = e^x \cos y, \quad v_y = e^x \cos y \implies u_x = v_y \quad \checkmark \] \[ u_y = -e^x \sin y, \quad v_x = e^x \sin y \implies u_y = -v_x \quad \checkmark \] Both Cauchy-Riemann equations are satisfied everywhere on $\mathbb{C}$.

Step 4: Final Answer
The function $e^x \cdot e^{iy} = e^z$ satisfies the Cauchy-Riemann equations. Thus, Option (D) is correct.