I think there might be an error in the official solution to the following question:
CMI May 23, 2022 BSc entrance exam, Question A2(7):
You are asked to take three distinct points $1, \omega_1, \omega_2$ in the complex plane such that $|\omega_1|=|\omega_2|=1$. Consider the triangle $T$ formed by the the complex numbers $1, \omega_1, \omega_2$:
(7) If $\omega_1+\omega_2$ is real, the triangle $T$ must be isosceles.
The official solution tells that the statement is true.
But I think I might have a counterexample:
What if we choose $\omega_1=\omega$ and $\omega_2=-\omega$?
Doesn’t that satisfy the criteria without the triangle being isosceles?
Perhaps ‘nonzero real’ might have been a more appropriate choice in the question statement.
Yes are true because is w1 and w2 both have same modulus 1 and if sum of both is purely real so Im(w1)=-Im(w2)=m
And we know that
Re(w1)²+Im(w2)²=1 Re(w2)²+Im(w2)²=1 // Modulus
subtract both
Re(w1)²=Re(w2)² Re(w1)=±Re(w2)=k
w1=k+im w2=±k-im w3=1
So if you draw these you can see that only w2=k-im is valid
So yes you are correct