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I have the following function and I am trying to find if it is analytic and differentiable. I use cauchy-riemann to prove it.

$ f(x) = x^2 -x+y+i(y^2-5y-x)$

$u(x,y) = x^2-x+y$ $v(x,y) = y^2-5y-x$

$u_x = 2x-1$ $u_y = 1$ $v_x= -1$ $v_y= 2y-5$

As a result $u_y = -v_x \Rightarrow 1 = -(-1) \Rightarrow 1 = 1$ and $u_x \neq v_y\Rightarrow y = x+2$

I was wondering if we can say that there some regions that the function is differentiable or analytic.

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This function fails to satisfy the Cauchy-Riemann equations and is is therefore not complex-differentiable.

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    On a separate note, a subset $G\subseteq\mathbb{C}$ is open if for each $z\in G$ there is some r > 0 so that the open ball centered at $z$, B_r(z) = \{w\in\mathbb{C}| |w - z| < r\} is contained in $G$.2012-01-26
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$h(x,y)=U(x,y)+iV(x,y)$

If $\partial_{x}(U(x,y))=\partial_{y}(V(x,y))$ and $\partial_{y}(U(x,y))=-\partial_{x}(V(x,y))$ then The function can be expressed as $h(x,y)=U(x,y)+iV(x,y)=f(z)=f(x+iy)$

For example

$h(x,y)=e^{x}\cos(y)+ie^{x}\sin(y)$ then

$U(x,y)=e^{x}\cos(y)$

$V(x,y)=e^{x}\sin(y)$

$\partial_{x}(U(x,y))=e^{x}\cos(y)$

$\partial_{y}(U(x,y))=-e^{x}\sin(y)$

$\partial_{x}(V(x,y))=e^{x}\sin(y)$

$\partial_{y}(V(x,y))=e^{x}\cos(y)$

$\partial_{x}(U(x,y))=\partial_{y}(V(x,y))$ and $\partial_{y}(U(x,y))=-\partial_{x}(V(x,y))$

Thus $h(x,y)$ can be expressed as $h(x,y)=f(z)=f(x+iy)$

Really if we check $h(x,y)=e^{x}\cos(y)+ie^{x}\sin(y)=e^{x}(\cos(y)+i\sin(y))=e^{x}e^{iy}=e^{x+iy}=e^{z}$

$h(x,y)=f(z)=e^{z}$

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    Mathlover, put a whack (\\) in front of sines and cosines. I did this; notice how it improved the appearance of your calculation.2012-01-27