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Let $f(z)=\frac{z^{5}}{|z|^4}$ for ($z\neq 0$) and $f(z)=0$ for $z=0$ then $u$ and $v$ satisfy Cauchy-Riemann equations everywhere but $f(z)$ is not differentiable at $z=0$ which means its not analytical because for a function to be analytic it must be differnetiable everywhere. I am confused because my teacher said that if $u$ and $v$ satisfy Cauchy-Riemann equations then $f(z)$ is analytic but this example proves that this is false. Am I misunderstanding something or is my teacher wrong?

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    Satisfying Cauchy-Riemann only is not enough. One needs more assumptions. One classical example is $f(z)=\exp(-1/z^4)$ for $z \neq 0$ and $f(0)=0$. See more at: http://www.jstor.org/stable/2321164?seq=1#page_scan_tab_contents – 2017-01-08
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    @Jack, what if they satisfy Cauchy-Riemann equations and satisfy Laplacian equation $u_{xx}+u_{yy}=0$ and $v_{xx}+v_{yy}=0$. Is $f$ analytical? – 2017-01-08
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    Harmonic functions are smooth. If $u$ and $v$ satisfy the C-R equations *and* are both $C^1$, then $f$ is analytic. – 2017-01-08
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    The $f$ in the question **does not** satisfy the CR equations at any point $z\ne0$. – 2017-01-08
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    The function given by $f(z) = z^5/|z|^4$ for $z ≠ 0$, $f(0) = 0$ is continuous everywhere and satisfies the Cauchy–Riemann equations at $z = 0$, but is not analytic at $z = 0$ (or anywhere else). – 2017-01-08
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    Note that holomorphic and complex differentiable are not synonyms. "Complex/real differentiable at $x$" just requires the limit of the difference quotient to exist at $x$; "holomorphic" requires it to exist everywhere in a neighborhood of $x$. The reason for this is that most of the nice theorems of complex analysis require holomorphicity. When the partial derivatives exist and are continuous, then you can say that $f$ is holomorphic in an open set $U$ iff it satisfies the CR equations. – 2017-01-09

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