Let $f(z)=\frac{1}{z}=u(x,y)+iv(x,y)$. Show that the circles of equations $u(x,y)=a$ and $v(x,y)=b$ ($a\not=0$,$b \not=0$, constants) are orthogonal curves.

One has $$\frac{1}{z} = \frac{1}{x+iy} = \frac{x}{x^2+y^2} +i \,\frac{-y}{x^2+y^2}.$$ Hence $u(x,y) = \frac{x}{x^2+y^2}$ and $v(x,y)=\frac{-y}{x^2+y^2}.$ The equation $u(x,y)=a$ gives $$ax^2+ay^2-x = 0.$$ Completing the square, I got $$\left(x-\frac{1}{2a}\right)^2+y^2 = \left(\frac{1}{2a}\right)^2.$$

The second equation is $x^2+(y+\frac{1}{2b})=(\frac{1}{2b})^2$

I dont know how to continue this problem. Any help?


  • $\begingroup$ The level curves of $u$ are orthogonal to the level curves of $v$, see here. $\endgroup$ – Ivo Terek Mar 6 '16 at 22:24
  • $\begingroup$ @IvoTerek So it is necessary to parameter each of the circles? $\endgroup$ – user316765 Mar 6 '16 at 22:26
  • 1
    $\begingroup$ No. Why would it be? We need only notice that $\nabla u$ and $\nabla v$ are orthogonal. $\endgroup$ – Ivo Terek Mar 6 '16 at 22:32
  • $\begingroup$ As I understand, since $u$ and $v$ are constants, so $<\nabla u,\nabla v>=0$? Why does it gives the informations $f(z)=\frac{1}{z}$? $\endgroup$ – user316765 Mar 6 '16 at 22:42
  • $\begingroup$ They give you a concrete expression for $f$ because probably they expect you to carry the computations for this case, not realizing that the result I mentioned above is true. $u$ and $v$ are not constants, $\langle \nabla u, \nabla v\rangle = 0$ follow from the Cauchy-Riemann equations $\endgroup$ – Ivo Terek Mar 6 '16 at 22:44

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