The question is as follows:
if $|a|=|b|=|c|=|b+c-a|=1$ where $a$,$b$,$c$ are distinct complex numbers , find $|b+c|$.
My attempt: By observation that $b=i$, $c=-i$, $a=1$ satisfy the following conditions
thus $|b+c| =|i-i|=0$
I realise that this is not a generalised method. Is there a better metod to solve this problem? Any help will be appreciated.
There are infinitely many solutions that satisfy that conditions.
Some of these will be in the form $b=-c$ and $|a|=1, a\in\Bbb C$. In this case, $|b+c|=0$ as you have discovered.
However, many more can be found. Define \begin{align} a&=\cos\alpha+i\sin\alpha\\ b&=\cos\beta+i\sin\beta\\ c&=\cos\gamma+i\sin\gamma \end{align} In order to satisfy $|b+c-a|=1$, we can have any set of $\alpha,\beta,\gamma$ such that $$\sqrt{(\cos\beta+\cos\gamma-\cos\alpha)^2+(\sin\beta+\sin\gamma-\sin\alpha)^2}=1.$$ Sure, it may be possible to parameterize your solution, but I suspect that there is more to the problem that has been left out. This is probably as far as we can go.