An analytic function f(z) of complex variable z = x + I y may be written as f(z) = u(x, y) + iv (x, y). Then, u(x, y) and v(x, y) must satisfy

The correct option/answer is the standard set of Cauchy-Riemann equations, which states that for a complex function to be analytic, its real and imaginary parts must satisfy:

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\text{and}\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

Here is the detailed step-by-step mathematical derivation of these equations based on the definition of analyticity:

Step 1: Definition of the Complex Derivative
For a complex function $f\left(z\right)=u(x,y)+iv(x,y)$ of the complex variable $z=x+iy$ to be analytic at a point, its derivative $f^{\prime }\left(z\right)$ must exist and have a unique value regardless of the path along which the limit is taken as $\Delta z\to 0$:

$$f^{\prime }\left(z\right)=\underset{\Delta z\to 0}{\lim }\frac{f(z+\Delta z)-f\left(z\right)}{\Delta z}$$

Step 2: Approaching the Limit along Path 1 (Horizontal Path)
If we approach the point along the real axis (horizontally), the change in the imaginary coordinate is zero, meaning $\Delta y=0$ and $\Delta z=\Delta x$:

$$f^{\prime }\left(z\right)=\underset{\Delta x\to 0}{\lim }\frac{\left[u(x+\Delta x,y)+iv(x+\Delta x,y)\right]-\left[u(x,y)+iv(x,y)\right]}{\Delta x}$$
Splitting this limit into real and imaginary components yields:

$$f^{\prime }\left(z\right)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$$

Step 3: Approaching the Limit along Path 2 (Vertical Path)
If we approach the point along the imaginary axis (vertically), the change in the real coordinate is zero, meaning $\Delta x=0$ and $\Delta z=i\Delta y$:

$$f^{\prime }\left(z\right)=\underset{\Delta y\to 0}{\lim }\frac{\left[u(x,y+\Delta y)+iv(x,y+\Delta y)\right]-\left[u(x,y)+iv(x,y)\right]}{i\Delta y}$$
Since $\frac{1}{i}=-i$, we can rewrite this expression as:

$$f^{\prime }\left(z\right)=\frac{1}{i}\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}$$

Step 4: Equating Real and Imaginary Parts
For the derivative to be unique and well-defined, the two limits obtained along Path 1 and Path 2 must be equal:

$$\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}$$
Equating the real parts on both sides gives:

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$
Equating the imaginary parts on both sides gives:

$$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}\text{which can be rearranged to}\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$