prove that there are no reals such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{x+y}$ I'm going through an introduction to complex analysis and there are two problems
that I'm having problems with.
A) Prove that there are no reals $x$ and $y$ such that $\displaystyle\frac{1}{x} + \frac{1}{y} = 
\frac{1}{x+y}$.
B) Prove that there are complex numbers $z$ and $w$ such that $\displaystyle\frac{1}{z} + \frac{1}{w} = \frac{1}{z+w}$.
Note that this is not homework. I'm just trying to get a better understanding of how to solve these types of problems. Thanks.
 A: HINT:
$(A)\implies (x+y)^2=xy\iff x^2+xy+y^2=0$
If $y=0,x^2=0\iff x=0$ and vice versa
Else $xy\ne0\implies \left(\dfrac xy\right)^2+\left(\dfrac xy\right)+1=0$
What about the discriminant of the above Quadratic Equation ? 
A: A somewhat (though perhaps only slightly) different approach:
I assume $x, y, x + y \ne 0$, so that the division operations are all well-defined.  Then the equation 
$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{x + y} \tag{1}$
becomes, upon multiplication by $x + y$, 
$\dfrac{x + y}{x} + \dfrac{x + y}{y} = 1, \tag{2}$
or, after a little turning of the algebra crank,
$2 + \dfrac{x}{y} + \dfrac{y}{x} = 1, \tag{3}$
or
$\dfrac{x}{y} + \dfrac{y}{x} + 1 = 0. \tag{4}$
Now set 
$\omega = \dfrac{x}{y} \tag{5}$
so that (4) becomes
$\omega + \omega^{-1} + 1 = 0 \tag{6}$
or
$\omega^2 + \omega + 1 = 0. \tag{7}$
The roots of (7) are
$\omega = \dfrac{1}{2}(-1 \pm i\sqrt 3); \tag{8}$
by (5), $x = \omega y$, so $x$ and $y$ cannot both be real.  Thus is item (A) resolved.  As for item (B), let $0 \ne z \in \Bbb C$ and set $w = \omega z$.  Then from (6),
$\dfrac{w}{z} + \dfrac{z}{w} + 1 = 0 \tag{9}$
or
$\dfrac{w}{z} + \dfrac{z}{w} + 2 = 1 \tag{10}$
or
$(\dfrac{w}{z} + 1) + (\dfrac{z}{w} +1) = 1 \tag{11}$
or
$\dfrac{w + z}{z} + \dfrac{w + z}{w} = 1. \tag{12}$
Note that
$w + z = \omega z + z = (1 + \omega)z \ne 0, \tag{13}$
since $1 + \omega \ne 0$; we can thus divide (12) by $w + z$, yielding
$\dfrac{1}{z} + \dfrac{1}{w} = \dfrac{1}{z + w}. \tag{14}$
And that takes care of item (B).  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
