1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.