# For which $(a,b) \in \mathbb{R}$ is $x^2+2axy+by^2$ the real part of a holomorphic function?

I am working on the following exercise:

For which $$a,b \in \mathbb{R}$$ is $$u := x^2+2axy+by^2$$ the real part of a holomorphic function in $$\mathbb{C}$$? For each of these $$(a,b)$$ find all holomorphic functions.

I think we should use the Cauchy-Riemann Differential Equations here, so for the imaginary part $$v$$ of a holomorphic function has to hold:

$$\frac{\partial u}{\partial_x} = 2x+2ay = \frac{\partial v}{\partial_y}$$

$$\frac{\partial u}{\partial_y} = 2ax+2by = -\frac{\partial v}{\partial_x}$$

To get $$v$$ from the partial derivatives I would say that we need integration. Integrating $$2x+2ay$$ over $$y$$ yields $$2xy+ay^2+C_1$$ and integrating $$(-1) \cdot (2ax+2by)$$ over $$x$$ yields $$-ax^2-2bxy+C_2$$ I do not know how to continue from here. Could you help me?

Assume that $$u$$ is the real part of a holomorphic function. By equality of mixed partial derivatives and the CR equations, $$\frac{\partial ^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial}{\partial y} \frac{\partial v}{\partial x} = 0$$

Holds for all $$(x,y) \in \mathbb C^2$$. This gives you $$b = -1$$.

We have : $$\frac{\partial v}{\partial x} = 2y - 2ax \implies v = f(y) + 2yx - ax^2$$ for some $$C^1$$ function of $$y$$. Similarly, $$\frac{\partial v}{\partial y} = 2x + 2ay$$ implies $$v = f(x) + 2xy + ay^2$$ for some $$C^1$$ function of $$x$$. Comparing the expression by isolating $$x$$ and $$y$$ terms, $$f(x) = -ax^2$$ and $$f(y) = ay^2$$. Thus, we get $$v(x,y) = -ax^2 + 2xy + ay^2$$.

Thus, $$x^2 + 2axy - y^2 + i(-ax^2+ 2xy + ay^2)$$ is the family of holomorphic functions in question.

Dig a little deeper, and you see that : $$x^2(1-ai) +2(a+i)(xy) - y^2(1-ai) = (1-ai)(x^2 + 2ixy - y^2) = (1-ai)(x+iy)^2$$ is a scalar multiple of the square of a complex number, so of course it is holomorphic.

If $$u$$ is the real part of a holomorphic function, $$u$$ is harmonic (and this follows from Cauchy-Riemann equations).

Thus, a necessary condition for $$y$$ to be the real part of a holomorphic function is that $$\Delta u=0\\ 2+2b=0\\ b=-1$$

Conversely, a harmonic function in all of $$\mathbb{R}^2$$ is the real part of an holomorphic function (which in our case you can determine explicitly), and we are done

The constants $$C_1=f(x)$$ and $$C_2=g(y)$$. Hence $$v_1=2xy+ay^2+f(x)$$ Signe we know $$v_{1x}=2y+f’(x)=-2ax-2by$$ From Which we get $$b=-1$$ and $$f’(x)=-2ax$$. We also have $$-ax^2-2byx+g(y)=v_2$$ We know that $$-2bx+g’(y)=2x+2ay$$ From comparison we get $$g’(y)=2ay$$. The two functions are identical so it doesn’t matter which one we choose. $$v_1=2xy+ay^2-ax^2$$ $$v_2=2xy+ay^2-ax^2$$ From these calculation I conclude that $$b=-1$$ and $$a$$ can be whatever.