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It is well-known that one of the form of the general solution of $u_{xx}+u_{yy}=0$ is $u(x,y)=C_1(x+iy)+C_2(x-iy)$ , where $C_1$ and $C_2$ are arbitrary functions.

However it is not quite convenient to have complex functions. Can we find another form of the general solution of $u_{xx}+u_{yy}=0$ : $u(x,y)=C_1(f(x,y))+C_2(g(x,y))$ , where $f(x,y)$ and $g(x,y)$ are real functions of $x$ and $y$ ?

Looking back to the second order linear ODE with constant coefficients $ay''+by'+cy=0$ , when $b^2-4ac<0$ , besides it has the form of the general solution $y=C_1e^{\frac{-b+\sqrt{b^2-4ac}}{2a}x}+C_2e^{\frac{-b-\sqrt{b^2-4ac}}{2a}x}$ , it also has the better form of the general solution $y=C_1e^{\frac{-bx}{2a}}\sin\dfrac{\sqrt{4ac-b^2}}{2a}x+C_2e^{\frac{-bx}{2a}}\cos\dfrac{\sqrt{4ac-b^2}}{2a}x$ so that we can avoid the complex function. Does the form of the general solution of $u_{xx}+u_{yy}=0$ have the similar trick?

Note that $f(x,y)$ and $g(x,y)$ I required can be any type of real functions and are not restricted to real polynomial functions only. In fact $u(x,y)=C_1(x+iy)+C_2(x-iy)$ can rewrite to $u(x,y)=C_1(e^{x+iy})+C_2(e^{x-iy})$ . So can $u(x,y)=C_1(e^x\sin y)+C_2(e^x\cos y)$ be the one of the form of the general solution of $u_{xx}+u_{yy}=0$ ?

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Regarding the edit: $(e^x\sin y)^2$ is not harmonic (indeed it has interior points of minimum where the sin vanishes), which was to be expected, given the answer by @RobertIsrael. – user31373 Sep 23 '12 at 3:28

Where $C_1$ and $C_2$ are again arbitrary functions? No.

If $f(x,y)$ is a $C^2$ real-valued function on ${\mathbb R}^2$ such that $C(f(x,y))$ is harmonic for arbitrary $C^2$ functions $C$ (or even the two specific functions $C(t) = t$ and $C(t) = t^2$), we have

$$ \left(\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2}\right) C(f(x,y)) = C''(f(x,y)) \left( \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 \right) + C'(f(x,y)) \left( \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \right) = 0 $$

Then both $\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2$ and $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ must be $0$. But $\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 = 0$ implies both $\partial f/\partial x$ and $\partial f/\partial y$ are $0$, and thus $f$ is constant.

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Complex numbers allow us to factor the sum of squares: $$\Delta=\left(\frac{\partial}{\partial x}\right)^2+\left(\frac{\partial}{\partial y}\right)^2 = \left(\frac{\partial }{\partial x}+i\frac{\partial }{\partial y}\right)\left(\frac{\partial }{\partial x}-i\frac{\partial }{\partial y}\right)$$ and, consequently, to obtain solutions of $\Delta u=0$ as the sum of solutions of $\left(\frac{\partial }{\partial x}+i\frac{\partial }{\partial y}\right)u=0$ and $\left(\frac{\partial }{\partial x}-i\frac{\partial }{\partial y}\right)u=0$. It's a fact of life that one cannot factor the polynomial $a^2+b^2$ over reals; complex numbers had to be invented for (approximately) this purpose.

To appreciate the benefit given by complex numbers in 2 dimensions, compare the above with what we have in three or higher dimensions, where the Laplacian cannot be factored without introducing some noncommutative weirdness.

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