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The problem is this:

Express $x$ and $y$ with $u$ and $v$, if $\dfrac{1}{x+iy} + \dfrac{1}{u+iv} = 1$

Where $x,y,u,v \in \mathbb{R}$, and $i^2 = -1$.

I could solve it, but I used a hairy and extremely large method: simplify the expression using Algebra, separate the real and imaginary part of it, and then use the substitution method recursively to get $x$ and $y$ as the problem states. As you may think now, I did a lot of simplifications and operations!

Is there another method to solve this?

Any help will be appreciated. Thank you! ;)

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In complex notation, we face $1/z+1/w=1$ to solve for $w$ we isolate it $1/w = 1-1/z$ then invert both sides to obtain $w = \frac{1}{1-1/z}$. Then, to find the solutions for $u,v$ in terms of $x,y$ I bet you can do.

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  • $\begingroup$ @jamesscook I see... But I have a question: this method works if I want $x$ and $y$ in terms of $u$ and $v$ (i.e. all the way around)? $\endgroup$ – user170247 Aug 18 '14 at 1:38
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    $\begingroup$ @TX286: Your expression is symmetric under $(u,v)\leftrightarrow(x,y)$. So if you solve for the former in terms of the latter, you'll get the reverse for free. $\endgroup$ – Semiclassical Aug 18 '14 at 1:59
  • $\begingroup$ @Semiclassical Oh I see... Let me try this, thank you! ;) $\endgroup$ – user170247 Aug 18 '14 at 2:05
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    $\begingroup$ @Semiclassical indeed, of course at the level of complex notation this is the $z \leftrightarrow w$ symmetry... $\endgroup$ – James S. Cook Aug 18 '14 at 5:36
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Hint

You can use brute force and rewrite you expression as $$(u+i v)+(x+i y)=(u+i v)(x+i y)$$ and develop. The real and imaginary parts give you two equations $$u+x-u x+v y=0$$ $$v- vx+y-u y=0$$ which are linear. Solving them is not too hard and you get $$x=\frac{u^2+v^2-u}{(u-1)^2+v^2}$$ $$y=-\frac{v}{(u-1)^2+v^2}$$

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  • $\begingroup$ Interesting. I'll check it out also! $\endgroup$ – user170247 Aug 18 '14 at 14:35

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