The prompt is:

"Suppose $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$ (Cauchy-Riemann equations.) Show that $\Re \left(\frac{1}{u + iv}\right)$ and $\Im \left(\frac{1}{u + iv}\right)$ satisfy the CR equations."

I understand Cauchy-Riemann, I understand all of the notation, and still I have no idea what this question means. The hint given is:

"The idea of [this problem] is that Cauchy Riemann for u and v make the Re and Im part of 1/(u+iv) also satisfy the CR equations." This sentence makes no sense to me. He continues: "This corresponds to the idea that if u+iv is complex differentiable then so is 1/(u+iv) by the calculus argument. So the CR equations for the 1 over function ought to follow from the CR equations for u and v--and this really happens!"

None of that makes sense to me.

What I don't get is how I'm supposed to use these two expressions. Is it $u = \Re \left(\frac{1}{u + iv}\right)$, $v = \Im \left(\frac{1}{u + iv}\right)$? That would make recursions though, which would make no sense. Is it $f = \Re \left(\frac{1}{u + iv}\right) + i\Im \left(\frac{1}{u + iv}\right)$? That would make slightly more sense but then why didn't he just write $f = \frac{1}{u + iv}$?

First I tried taking $\frac{\partial}{\partial x}$ of the real part, which I'll call $\Re$. I got a large expression with $u$'s and $v$'s and partial derivatives mixed around in it. Then I took the $\frac{\partial \Re}{\partial y}$ of the same thing and I discovered that $\frac{\partial \Re}{\partial y} \not = \frac{\partial \Re}{\partial x}$, $\frac{\partial \Re}{\partial y} \not = -\frac{\partial \Re}{\partial x}$, $\frac{\partial \Re}{\partial y} \not = -i\frac{\partial \Re}{\partial x}$. I'm stabbing in the dark here but none of these seem to satisfy any CR equation.

Then I figured he's using $u$ and $v$ where he shouldn't be, because $f = u + iv$ not $f = \frac{1}{u + iv}$. So I made a new equation, $f(x + iy) = \frac{1}{a(x + iy) + ib(x + iy)}$. This is really just $f(z) = z^{-1}$ so it should satisfy CR and be complex differentiable. Maybe that's all he's trying to get at. So I take $\frac{\partial f}{\partial x} = \frac{\frac{\partial a}{\partial c} + i\frac{\partial b}{\partial c}}{(a+ib)^2}$ and $\frac{\partial f}{\partial y} = \frac{\frac{\partial a}{\partial y} + i\frac{\partial b}{\partial y}}{(a+ib)^2}$. If he's supposing that the CR equations hold (that's his hypothesis) then $\frac{\partial f}{\partial x} = -i\frac{\partial f}{\partial y}$. Plugging those in, the denominators cancel and I get $\frac{\partial a}{\partial x} + i\frac{\partial b}{\partial x} = -i(\frac{\partial a}{\partial y} + i\frac{\partial b}{\partial y}) = -i\frac{\partial a}{\partial y} + \frac{\partial b}{\partial y}$, so matching up the real and imaginary parts gets me $\frac{\partial b}{\partial x} = -\frac{\partial a}{\partial y}$ and $\frac{\partial a}{\partial x} = \frac{\partial b}{\partial x}$. Well that's CR I suppose, do I win? But I don't feel like I learned anything from the experience, and I never used $\Re \left(\frac{1}{u + iv}\right)$ and $\Im \left(\frac{1}{u + iv}\right)$ so it can't be right.

Final note: Yes this is my homework, but I explicitly do not want the answer, I would just like an explanation of the meaning of the question so I can figure out the answer myself.


$$\frac1{u+iv}=\frac{u}{u^2+v^2}-\frac{v}{u^2+v^2}i.$$ I.e., $$\Re\left(\frac{1}{u+iv}\right)=\cdots$$ $$\Im\left(\frac{1}{u+ v}\right)=\cdots$$

  • $\begingroup$ $$\Re\left(\frac{1}{u+iv}\right)=\frac{u}{u^2+v^2}$$ $$\Im\left(\frac{1}{u+iv}\right)=\frac{-v}{u^2+v^2}$$ I'm not sure I get your point? $\endgroup$ – Jorge Rodriguez Oct 16 '14 at 16:07
  • $\begingroup$ Check if this pair of functions verifies CR equations. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 17 '14 at 9:58
  • $\begingroup$ Okay if I ignore the recursion of $u=\frac{u}{u^2+v^2}$ then it does seem to work. Thanks. $\endgroup$ – Jorge Rodriguez Oct 18 '14 at 15:37
  • $\begingroup$ No recursion at all. $u$, $v$ are simply names. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 18 '14 at 20:08

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