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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!

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  • $\begingroup$ Can you see the left hand side of the equation you got is identically $0$? You also need to check the other Cauchy-Reimann equation. I vote to close this question, because it seems too localized to me. $\endgroup$ – 23rd Apr 29 '13 at 20:17
  • $\begingroup$ Oh wow, I feel so dumb now, it does equal 0, I'm really sorry. $\endgroup$ – aNxello Apr 29 '13 at 20:23
  • $\begingroup$ Never mind. By the way, note that $u+iv=\frac{z(1+\bar{z})}{(1+z)(1+\bar{z})}=\frac{z}{1+z}$, so it is holomorphic(complex analytic). $\endgroup$ – 23rd Apr 29 '13 at 20:24
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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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  • $\begingroup$ I hadn't realized the first equation did hold, I will now do the second, thank you very much. $\endgroup$ – aNxello Apr 29 '13 at 20:23
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Observe $-2x^2-4x-2=-2(x^2+2x+1) = -2(x+1)^2$.$$x(2(1+x)^2-2x^2-4x-2)=x(2(1+x)^2-2(1+x)^2)=0x=0$$

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