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I want to check if there is (or not) an analytic function on $\mathbb{C}$ \ ${0}$ such that $Im(f)=78+x^2-5y^6- \frac{5y}{3(x^2+y^3)}$. What I thought of doing is first applying the Identity Theorem which guarantees the uniqueness of the analytic function (if it really exists), so I just set $x=z$ and let $y=0$. So I would get $f(z)=u(z,0)+iv(z,0)$.

If $f$ is analytic then by the Cauchy-Riemann equations, we get $v_x=2x=-u_y$ and $v_y=0=u_x$. So $u$ does not depend on $x$, which contradicts that $u_y=-2x$. So there is no such analytic function by the Identity Theorem.

Is it that easy or did I just do complete nonsense by first applying the Identity Theorem then going to the partial derivatives? Can I do that? Thx.

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  • $\begingroup$ What you did would "work" for ANY function, which should tell you that your reasoning is incorrect. What is the "Identity Theorem" that you refer to? $\endgroup$
    – Corey
    Jul 2, 2011 at 5:21
  • $\begingroup$ See connections with complex function theory in en.wikipedia.org/wiki/Harmonic_function $\endgroup$
    – jspecter
    Jul 2, 2011 at 5:27

2 Answers 2

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Generally, you can test whether or not a function of two variables can be considered a real/imaginary part by whether or not it is harmonic, i.e. it is in fact (locally) part of an analytic function if and only if $ \Delta v = v_{xx} + v_{yy} = 0 $. If you happen to know for a fact that $ v $ is harmonic, then you can solve for $ u $ with Milne's method: Write $ f = u + i v$, then $ f' = u_x + i v_x $, and by Cauchy-Riemann we can substitute $ v_y $ for $ u_x$, yielding $ f' = v_y + i v_x $; write the latter expression as a function of the complex variable $ z = x+iy $ in order to find $ f'(z) $ in a useful form; finally integrate to obtain $ f(z) $ up to a constant and then subtract out the imaginary part $ i v $ in order to obtain the real part $ u $ left over.

In this case $ \Delta \left( 78 +x^2 - 5y^2 - \frac{5y}{x^2+y^2} \right) = -8 \ne 0 $, so your function cannot be the real or imaginary part of any analytic function. Note that you can't hold $ y $ fixed while taking a partial derivative with respect to $ y $ so your method of deduction is invalid.

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  • $\begingroup$ Thank you so much, that was really clear. $\endgroup$
    – user786
    Jul 2, 2011 at 5:45
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Cauchy-Riemann is the right idea (but the wrong execution). You're given $v(x,y)$, you can calculate $v_x$ and $v_y$, then use Cauchy-Riemann to find $u$ or to prove it doesn't exist.

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  • $\begingroup$ Actually Cauchy-Riemann doesn't say explicitly whether or not $ u $ exists or what form it takes - only what its partials would be if it does exist. $\endgroup$
    – anon
    Jul 2, 2011 at 5:41
  • $\begingroup$ I did that but I found a pretty nasty expression which I need to integrate to find $u$ $\endgroup$
    – user786
    Jul 2, 2011 at 5:44

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