For this question it is assumed that $f(x+iy)$ is differentiable on $C$. For a function $f(x+iy)=u(x,y)+iv(x,y)$, the real part $u(x,y)=x^3y-xy^3$. Now, I am trying to find the imaginary part of the function, so I used the Cauchy Riemann equations and I got $\frac{\partial u}{\partial x}=3x^2y-y^3=\frac{\partial v}{\partial y}$ and $-\frac{\partial u}{\partial y}=3xy^2-x^3=\frac{\partial v}{\partial x}$.
Finding imaginary part of a function $f(x+iy)=u(x,y)+iv(x,y)$
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2Your constant depends on $x$ in the first case... – 2011-11-10
2 Answers
Your very close to the correct answer. When you compute the integrals the 'constants' are constant with respect to the variable of integration. For example, your first integral should be:
$v=\int 3x^2y-y^3 dy = \frac{3}{2}x^2y^2-\frac14y^4 + \phi(x)$
where $\phi(x)$ is a function of $x$ to be determined.
Now differentiating with respect to $x$, then using the Cauchy-Riemann equations gives:
\frac{\partial v}{\partial x} = 3xy^2+\phi'(x) \quad \Rightarrow \quad \phi'(x) = -\frac{\partial u}{\partial y} - 3xy^2 = -x^3
So that $\phi(x) = -\frac14 x^4+C$, for constant $C$. Thus we have: $v=\frac{3}{2}x^2y^2-\frac14y^4-\frac14x^4 + C$
Notice that this means that we can only determine $f$ up to a constant.
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0How would we know that no such function $v(x,y)$ exists? – 2013-09-17
A non-systematic answer: One immediately sees that $f$ must be purely imaginary on both the real and the imaginary axis. Also, the real part at least is purely a fourth degree polynomial of in the components of $z$. Together this suggests a function of the form $f(z)=ciz^4$ for some real $c$ ...
(One lesson to take home here, apart from serendipity, is if one of the coordinate functions for a differentiable complex function has a nice algebraic expressions, then there is almost always also an nice expression for the full function that uses purely complex operations. Searching for that form can give a much more satisfying answer than one phrased purely in terms of $x$s and $y$s).