Question

please answer these four
questions

Answer 1

z₁ = r cis θ and z₂ = r cis(θ+π/3) for an arbitrary equilateral triangle, where r>0, then...

z₁² + z₂² - z₁z₂
= r² cis(2θ) + r² cis(2θ+2π/3) - r²cis(2θ+π/3)
= r²( cis(2θ) + cis(2θ+2π/3) - cis(2θ+π/3) )
= r²( ( cos(2θ) + cos(2θ+2π/3) - cos(2θ+π/3) ) + i( sin(2θ) + sin(2θ+2π/3) - sin(2θ+π/3) ) )
= r²( ( cos(2θ) + cos(2θ)cos(2π/3) + sin(2θ)sin(2π/3) - cos(2θ)cos(π/3) - sin(2θ)sin(π/3) ) + i ( sin(2θ) + sin(2θ)cos(2π/3) + cos(2θ)sin(2π/3) - sin(2θ)cos(π/3) - cos(2θ)sin(π/3) ) )
= r²( ( cos(2θ)(1 + cos(2π/3) - cos(π/3)) + sin(2θ)(sin(2π/3) - sin(π/3) ) + i ( sin(2θ)(1 + cos(2π/3) - cos(π/3))+ cos(2θ)(sin(2π/3) - sin(π/3) ) ) )
= r²( ( cos(2θ)(0) + sin(2θ)(0) ) + i( sin(2θ)(0)+ cos(2θ)(0) ) )
= r²( 0+0i)
= 0
Thanks