If 1/(a+b) = 1/a + 1/b , what is the solution for z=a/b? [closed]

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If 1/(a+b) = 1/a + 1/b , what is the solution for z=a/b?

I made an attempt to use the formula for the solution of an equation of a complex number but I don't know how to use it in this case.

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3 Answers

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Here are equivalent formulations $$\begin{eqnarray*} \frac{1}{a+b} &=& \frac{1}{a} + \frac{1}{b}\\ 1 &=& \frac{a+b}{a} + \frac{a+b}{b}\\ 1 &=& 2 + \frac{b}{a} + \frac{a}{b}\\ 0 &=& 1 + \frac{1}{z} + z\\ 0 &=& z^2 + z+ 1\\ 0 &=& \left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\\ z &=& -\frac{1}{2} \pm \frac{i}{2} \sqrt 3 \end{eqnarray*}$$

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HINT:if $$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$$ then we have $$ab=(a+b)^2$$ or $$0=a^2+ab+b^2$$ this can be written as $$\frac{a}{b}+1+\frac{b}{a}=0$$ thus when we set $$t=\frac{a}{b}$$ then we get? we also Need $$a,b\ne 0$$ and $$a+b\ne 0$$

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Just to be different:

$\frac 1a + \frac 1b = \frac 1{a+b}$

$(a+b)(\frac 1a + \frac 1b) = 1$

$1 + \frac ab + \frac ba + 1 = 1$

$1 + z + \frac 1z = 0$

$z^2 + z + 1 = 0$

$(z-1)(z^2 + z + 1) = z^3 - 1 = 0*(z-1)$

$z^3 = 1$ but clearly $z \ne 1$ as $1^2 + 1 + 1 \ne 0$.

$z$ a cube root of unity but not the real cube root of unity.

$z = e^{\pm \frac {2\pi}3i}= \cos(\pm \frac {2\pi}3) + i \sin(\pm \frac {2\pi}3) = -\frac 12 \pm i\frac{\sqrt{3}}2$

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Or to be really different:

$a =\frac 1{\frac 1{a+b}} - b$

$= \frac 1{\frac 1a + \frac 1b} - b$

$= \frac {ab}{a+b} -b$

So $a^2 + ba = ab - b^2 - ab$

$a^2 + ab + b^2 = 0$

So $a = \frac {-b \pm \sqrt{b^2 - 4b^2}}2$

$=\frac {-b \pm |b|\sqrt {-3}}2$

$= \frac { -b \pm b i\sqrt{3}}2$

So $\frac ab = \frac {\frac { -b \pm bi \sqrt {3}}2}b= \frac { -1 \pm i\sqrt {3}}2$

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