# Finding an analytic function given the real/imaginary part

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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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? –  Corey Jul 2 '11 at 5:21
See connections with complex function theory in en.wikipedia.org/wiki/Harmonic_function –  jspecter Jul 2 '11 at 5:27

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.
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.
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. –  anon Jul 2 '11 at 5:41
I did that but I found a pretty nasty expression which I need to integrate to find $u$ –  user786 Jul 2 '11 at 5:44