<s>[INST] $f(z)=\frac{1}{z}=u(x,y)+iv(x,y)$ - Circles of equations $u(x,y)=a$ and $v(x,y)=b$ are orthogonal curves. 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.
I am stuck for a while on this problem. How could do this question?
 [/INST] 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, you get $$\left(x-\frac{1}{2a}\right)^2+y^2 = \left(\frac{1}{2a}\right)^2.$$ Do the same with the other circle and you should get the conclusion.
</s>