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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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2 Answers 2

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

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I was wondering if we can say that is differentiable in some specific region. So that means it is not analytic too. I guess I guess didn't understand what open region means. –  primer Jan 26 '12 at 15:25
This is an example of a function that is differentiable as a map from the plane into itself as a real (vector) function. However it is not analytic –  ncmathsadist Jan 26 '12 at 15:37
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 $G$. –  ncmathsadist Jan 26 '12 at 15:39


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







$\partial_{x}(U(x,y))=\partial_{y}(V(x,y))$ 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 $h(x,y)=e^{x}\cos(y)+ie^{x}\sin(y)=e^{x}(\cos(y)+i\sin(y))=e^{x}e^{iy}=e^{x+iy}=e^{z}$


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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. –  ncmathsadist Jan 27 '12 at 2:09

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