What I have done so far isn't so helpful. If $z_1=x_1+y_1i$ and $z_2=x_2+y_2i$ then these two things must hold: $$x_1^2 -y_1^2 +x_2^2-y_2^2+x_1x_2-y_1y_2=0$$ $$2x_1y_1+2x_2y_2+x_1y_2+x_2y_1=0$$ This also means that $$\Re(z_1z_2)=y_1^2+y_2^2-(x_1^2+x_2^2)$$ $$\Im(z_1z_2)=-2(x_2y_2+x_1y_1)$$ But again, this isn't very helpful. Are there any simpler solutions that don't involve this nasty algebra?

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    $\begingroup$ Notice that multiplying both sides of your original equation by $z_1z_2$ gives $z_1+z_2=1$. (But don't let this fool you into thinking that $(1,0)$ or $(0,1)$ is a solution.) $\endgroup$ – 211792 Aug 31 '16 at 21:30
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    $\begingroup$ Haha I feel ashamed... Thank you. $\endgroup$ – darmendarizp Aug 31 '16 at 21:45
  • $\begingroup$ EDIT: Well, I got confused. The problem really was $\frac{1}{z_1}+\frac{1}{z_2}=\frac{1}{z_1+z_2}$ not $\frac{1}{z_1}+\frac{1}{z_2}=\frac{1}{z_1z_2}$ $\endgroup$ – darmendarizp Aug 31 '16 at 22:10
  • $\begingroup$ FYI, this question inspired me to create one of my own; see the Linked question in the list to the right. $\endgroup$ – Semiclassical Sep 1 '16 at 17:36

As long as $z_1,z_2,z_1+z_2\neq 0$, you can cross-multiply to get an equivalent equation $z_2(z_1+z_2)+z_1(z_1+z_2)=z_1z_2$, which rearranges to $$z_1^2+z_1z_2+z_2^2=0.$$ By the quadratic formula you can solve for $z_1$ in terms of $z_2$: $$z_1=\frac{-z_2\pm\sqrt{z_2^2-4z_2^2}}{2}=\frac{-1\pm i\sqrt{3}}{2}z_2.$$

So your original equation has all the solutions to this except for those for which $z_1$, $z_2$, or $z_1+z_2$ are $0$ (in fact, you can check that the only way any of them can be $0$ is if $z_1=z_2=0$).

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    $\begingroup$ To get the result without using the quadratic formula, one can multiply the quadratic by $z_1-z_2$. This obtains $z_1^3-z_2^3=0$ i.e. $z_1/z_2$ is a cubic root of unity. $\endgroup$ – Semiclassical Aug 31 '16 at 22:34

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