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    If f(z) = u+iv is analytic is f*(z) = u-iv analytic?

    I know that f*(z) is not analytic because the Cauchy-Riemann equations would not be satisfied but how would I show it?

    I'm a little confused because if it was, f(z)= x+iy and f*(z) = x-iy I would know how to show that f*(z) is not analytic using C-R equations because u(x,y) = x and v(x,y)=y and I'd just use the c-r equations and show they are not the same.
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    Um, it's analytic if f is constant...
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    I mean from the first one you know that the CR equations are satisfied right? Therefore, as V1(x,y) = -V2(x,y), then cannot be satisfied for the second equation because (partial V1)/(partial y) = -(partial V2)/(partial y) hence they cannot both equal (partial u)/(partial x).
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    (Original post by DFranklin)
    Um, it's analytic if f is constant...
    My bad didn't realise that. But what if f is not a constant?
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    if f is constant then the CR equations are simply 0=0 so it's a somewhat trivial solution.
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    (Original post by Darth_Narwhale)
    if f is constant then the CR equations are simply 0=0 so it's a somewhat trivial solution.
    Yeah, but look at it the other way: you're going to need f'(z) non-zero somewhere to be able to deduce a contradiction to the C-R equations. OP seems to have given up anyhow, however.
 
 
 
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Updated: February 13, 2017
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