<s>[INST] Find all solutions to $\frac{a+b}{b} = \frac{a}{a+b}$ I want to find all solutions to $\frac{a+b}{b} = \frac{a}{a+b}$. I have plugged this into a calculator which tells us that if we solve for $a$, without loss of generality, we obtain
$$a = - \frac{b}{2} \pm \frac{\sqrt{3}}{2}i$$
and vice versa. I am not sure how to show this, however. I have tried a couple of different things. For example,
Method 1:
$$\frac{a}{a+b} = \frac{a+b}{b} \implies (a+b)^2=ab \implies a+b = \pm \sqrt{ab}$$
Method 2:
$$\frac{a}{a+b} \cdot \frac{a-b}{a-b}= \frac{a+b}{b} \cdot \frac{a-b}{a-b} \implies \frac{a(a-b)}{a^2-b^2}=\frac{a^2-b^2}{b(a-b)} \implies \big( a^2-b^2 \big)^2 = ab(a-b)^2$$
But so far neither of these methods are shedding any light for me. Any advice? Thanks in advance.
 [/INST] You have $(a+b)^2=ab$ or $a^2+ab+b^2=0$. If $a\ne b$, multiply by $(a-b)$, get $a^3-b^3=0$. So $a=b\omega$ where $\omega$ is one of the cube roots of $1$. The possibility $a=b$ does not work, the other two work.
Answer: $b$ is any non-zero number, $a=b\omega$ where $\omega$ is one of the two non-1 cube roots of $1$.
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