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.

To begin with, let us assume that $b\neq 0$ and $a\neq -b$.
Now we can multiply both sides by $b(a+b)$ in order to obtain
\begin{align*}
(a + b)^{2} = ab & \Longleftrightarrow a^{2} + 2ab + b^{2} = ab\\\\
& \Longleftrightarrow a^{2} + ab + b^{2} = 0\\\\
& \Longleftrightarrow \left(a + \frac{b}{2}\right)^{2} + \frac{3b^{2}}{4} = 0\\\\
& \Longleftrightarrow \left(a + \frac{b}{2}\right)^{2} = -\frac{3b^{2}}{4}\\\\
& \Longleftrightarrow a + \frac{b}{2} = \pm\frac{bi\sqrt{3}}{2}\\\\
& \Longleftrightarrow a = -\frac{b \pm bi\sqrt{3}}{2}
\end{align*}
Hopefully this helps !
