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 February 25th, 2014, 03:28 PM #1 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Show function is not analytic Please check my work on the following problems Show that the given function is not analytic at any point 1) $f(z)= Re(z)$ $Re(z)= x$ $\frac{\partial u}{\partial x}=1$ and $\frac{\partial v}{\partial y}= 0$, and $1 \neq 0$ 2) $f(z)= y + ix$ $\frac{\partial u}{\partial y}=1$ and $\frac{\partial v}{\partial y}= 0$, so $\frac {\partial u}{\partial y} \neq - \frac{\partial v}{\partial y}$ since $1 \neq 0$ Approaching this right? Thanks
 February 25th, 2014, 07:53 PM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Show function is not analytic Revise number (2) . Another way of doing it is $f=y+xi= i(x-iy) = i \bar{z}$ $\frac{df}{d\bar{z}}=i \neq 0$ So it is not analytic anywhere.

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