Getting two different answers for the imaginary part of an analytic function.

Question

I have been studying complex analysis (informally) for a while, and everything’s fine. But, I got stuck at a very simple problem today.

The book mainly want to show that, if the real part of an analytic function is given, its imaginary part can be determined uniquely up to a constant.

$f=u+iv$

Given $u=x^2-y^2$.

Then, $$dv=2(ydx+xdy)$$

Method 1:

Recognize that $$ydx+xdy=d(xy)$$, immediately $v=2xy+C$ which is the correct answer.

Method 2:

Since $x$ and $y$ are independent of each other, we have $$v=\int dv=2\int\left(ydx+xdy\right)=2\int ydx+2\int xdy=2y\int dx+2x\int dy=4xy+C$$ which is wrong.

What’s the problem?

Answer 1

When doing integration (and using it to find a primitive using FToC) its important to specify a domain. Your first attempt can be justified by considering the variable $w=xy$, which is then integrated along the real line. However, in the second attempt, you start off with a single integral, which you then split, and integrate one part along a curve of constant $x$ and varying $y$, and the other along a curve of constant $y$ and varying $x$, this us not permitted. If you want to use FToC for line integrals in high dimensional space then this is $$ \int_C \nabla f \cdot dl = f(x_1) - f(x_0) $$ where $x_0$ and $x_1$ are the start and end of the curve $C$ respectively.

Answer 2

That just goes to show what a hateful notation $$\int\left(f(x,y)dx+g(x,y)dy\right)$$ is for a line integral. A one-dimensional integral must have a single variable of integration, but you wouldn't know it from looking at the above expression. If you write it out as $$\int\vec F\left(\vec r\right)\cdot d\vec r$$ Then it's clear that we must specify some path in the $(x,y)$-plane, parameterize, perhaps piecewise, in terms of a single variable, then perform the integration. As an example, suppose we start at $(x,y)=(0,0)$, then integrate to $(x,0)$ along the $x$-axis, and finally to $(x,y)$ along a line of constant $x$. Along the first contour, $(x,y)=(s,0)$, $\vec F\left(\vec r\right)=\langle2y,2x\rangle=\langle0,2s\rangle$, and $d\vec r=d\langle s,0\rangle=\langle1,0\rangle ds$ so $$v(x,0)=v(0,0)+\int_0^s\langle0,2s\rangle\cdot\langle1,0\rangle ds=v(0,0)+\int_0^s0ds=v(0,0)$$ Along the second contour, $(x,y)=(x,t)$, $\vec F\left(\vec r\right)=\langle2t,2x\rangle$, and $d\vec r=\langle0,1\rangle dt$. Then $$v(x,y)=v(x,0)+\int_0^y\langle2t,2x\rangle\cdot\langle0,1\rangle dt=v(0,0)+\int_0^y2x\,dt=v(0,0)+2xy$$ So we arrive at the same answer we would have by saying that $$\frac{\partial v}{\partial x}=2y$$ so $$v=2xy+f(y)$$ And then $$\frac{\partial v}{\partial y}=2x+f^{\prime}(y)=2xy$$ So $f^{\prime}(y)=0$ and $f(y)=C$, so $v(x,y)=2xy+C$.