triangle (1 Viewer)

[jnney]

If the points A and B represent the numbers z1 and z2 respectively on an argand diagram, and triangle OAB is equilateral, prove that z1^2 + z2^2 =z1z2

 

[math man]

since OAB is equilateral z2 = cis60 z1...now this is a proof question so taking the LHS= z1^2 + z2^2 this becomes z1^2 + (cis60z1)^2...Simplyfying further...LHS= z1^2 + cis120z1^2=z1^2(1+cis120)=z1^2cis60
as if you expand 1 + cis120 you will get cis60... therefore LHS = z1(z1cis60) = z1 z2 = RHS

 

[Carrotsticks]

This proof derives the required identity from scratch, rather than just substituting in things into LHS, and manipulating it to make it become RHS.

This could just be me, but I don't feel right 'proving' an identity by doing this, since this means that you must have known the identity in the first place! I would much rather start from something known, then work my way up. I've attached it below. I cbb using latex right now, so I'll just use my trusty tablet.



I used the fact that z1^2 + z2^2 = z1z2 SORTA looks like the binomial expansion (z1+z2)^2 = z1^2 + 2z1z2 + z2^2........... as a starting point.

Remember to keep an eye out for little clues like these. Use the terms of the question as a clue as to what to do.