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I'm having a really tough time understanding complex analysis. The assignment is to find a function $f$, given $z = x + iy$, so that f is analytic on the complex plane, so that $\mathrm{Re}(f) = x^3 y - x y^3$.

I tried beginning with the Cauchy-Riemann method and found the partial derivatives for the function $u(x, y)$ but I have no idea how to get to the original function $f$.

Any help here?

Rhys
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dingari
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  • With simple equations and symbols, adding a dollar sign before and after is enough to have them render nicely. For more details, start here: http://math.stackexchange.com/help/notation – Rhys Sep 05 '13 at 11:12

1 Answers1

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Write $f = u+iv$, with real-valued functions $u,\, v$. The Cauchy-Riemann equations are

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}; \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}.$$

So with the given $u(x,y) = x^3y - xy^3$, the first yields

$$\frac{\partial u}{\partial x} = 3x^2y - y^3 = \frac{\partial v}{\partial y},$$

and thus $v(x,y) = \frac{3}{2}x^2y^2 - \frac{1}{4}y^4 + \varphi(x)$ with an unknown function $\varphi$. The second CR equation then yields

$$-\frac{\partial u}{\partial y} = 3xy^2 - x^3 = 3xy^2 + \varphi'(x),$$

so we obtain $\varphi(x) = -\frac14 x^4 + c$ with an unknown (real) constant $c$.

Assembling,

$$v(x,y) = - \frac14x^4 + \frac32 x^2y^2 - \frac14y^4 + c = -\frac14(x^4 - 6x^2y^2 + y^4) + c,$$

so

$$f(x+iy) = x^3y - xy^3 -\frac{i}{4}(x^4 - 6x^2y^2 + y^4) + ic = \frac{1}{4i}(x+iy)^4 + ic.$$

Daniel Fischer
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