1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $$\Re\left(\frac{1}{u+iv}\right)=\frac{u}{u^2+v^2}$$ $$\Im\left(\frac{1}{u+iv}\right)=\frac{-v}{u^2+v^2}$$ 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

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.