<s>[INST] Complex analytic function and Cauchy-Riemann conditions question here again asking for some guidance
Let me state the problem and working on and then my question,
So I have two functions:
$$u(x, y) = \frac{x(1+x)+y^2}{(1+x)^2 + y^2}$$
and
$$v(x, y) = \frac{y}{(1+x)^2 + y^2}$$
and I'm being asked if they can serve as real and imaginary parts respectively of some analytic function of $z = x + iy$
So what I thought I had to do was check for the Cauchy-Riemann conditions, so I started with the first condition
$$\frac{du( x,y )}{dx} = \frac{dv( x,y )}{dy}$$
and I got $x(2(1+x)^2 -2x^2 -4x -2) = 0$
does that mean they don't satisfy the equation? does it? should I check for the other condition?
Thank you!
 [/INST] The Cauchy-Riemann equations on a pair  functions $u(x,y)$ and $v(x,y)$ are the two equations:
$\displaystyle \frac{du}{dx} = \frac{dv}{dy}$ and $\frac{du}{dy} = -\frac{dv}{dx}$.
Checking these two equations for the given functions $u(x,y)$ and $v(x,y)$ gives that
$\displaystyle \frac{du}{dx} = \frac{1+ 2x + x^2 - y^2}{(1 + 2x + x^2 + y^2)^2}$ and
$\displaystyle \frac{dv}{dy} = \frac{1+ 2x + x^2 - y^2}{(1 + 2x + x^2 + y^2)^2}$ so the first equation holds. For the second one note that $\displaystyle \frac{du}{dy} = \frac{2(1+x)y}{(1 + 2x + x^2 + y^2)^2}$ and $\displaystyle \frac{dv}{dx} = -\frac{2(1+x)y}{(1 + 2x + x^2 + y^2)^2}$ so the second equation also holds. You can derive the appropriate conclusion yourself.
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