Criteria Marks Provides correct solution 3 Factors z 2 3 + z 1 3 z_2^3+z_1^3 z 2 3 + z 1 3 OR <br>Simplifies either side of z 1 2 + z 2 2 = z 1 z 2 z_1^2+z_2^2=z_1z_2 z 1 2 + z 2 2 = z 1 z 2 z 2 = e i π / 3 z 1 z_2=e^{i\pi/3}z_1 z 2 = e iπ /3 z 1 2 Observes that z 2 3 = e i π z 1 3 z_2^3=e^{i\pi}z_1^3 z 2 3 = e iπ z 1 3 OR <br>Uses z 2 = e i π / 3 z 1 z_2=e^{i\pi/3}z_1 z 2 = e iπ /3 z 1 z 1 2 + z 2 2 = z 1 z 2 z_1^2+z_2^2=z_1z_2 z 1 2 + z 2 2 = z 1 z 2 1
Sample answer:
z 2 = e i π / 3 z 1 z_2=e^{i\pi/3}z_1 z 2 = e iπ /3 z 1 ∴ z 2 3 = e i π z 1 3 = − z 1 3 \therefore z_2^3=e^{i\pi}z_1^3=-z_1^3 ∴ z 2 3 = e iπ z 1 3 = − z 1 3 ∴ z 1 3 + z 2 3 = 0 \therefore z_1^3+z_2^3=0 ∴ z 1 3 + z 2 3 = 0 ( z 1 + z 2 ) ( z 1 2 + z 2 2 − z 1 z 2 ) = 0 (z_1+z_2)(z_1^2+z_2^2-z_1z_2)=0 ( z 1 + z 2 ) ( z 1 2 + z 2 2 − z 1 z 2 ) = 0 Now z 1 ≠ − z 2 z_1\ne-z_2 z 1 = − z 2
z 1 2 + z 2 2 = z 1 z 2 z_1^2+z_2^2=z_1z_2 z 1 2 + z 2 2 = z 1 z 2 OR
LHS = z 1 2 + z 2 2 = ( e 2 π i / 3 + 1 ) z 1 2 \text{LHS}=z_1^2+z_2^2=(e^{2\pi i/3}+1)z_1^2 LHS = z 1 2 + z 2 2 = ( e 2 π i /3 + 1 ) z 1 2 = ( cos 2 π 3 + i sin 2 π 3 + 1 ) z 1 2 =\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}+1\right)z_1^2 = ( cos 3 2 π + i sin 3 2 π + 1 ) z 1 2 = ( 1 2 + i 3 2 ) z 1 2 =\left(\frac12+i\frac{\sqrt3}{2}\right)z_1^2 = ( 2 1 + i 2 3 ) z 1 2 RHS = z 1 z 2 = e i π / 3 z 1 2 = ( cos π 3 + i sin π 3 ) z 1 2 \text{RHS}=z_1z_2=e^{i\pi/3}z_1^2=\left(\cos\frac\pi3+i\sin\frac\pi3\right)z_1^2 RHS = z 1 z 2 = e iπ /3 z 1 2 = ( cos 3 π + i sin 3 π ) z 1 2 = ( 1 2 + i 3 2 ) z 1 2 =\left(\frac12+i\frac{\sqrt3}{2}\right)z_1^2 = ( 2 1 + i 2 3 ) z 1 2 Therefore z 1 2 + z 2 2 = z 1 z 2 z_1^2+z_2^2=z_1z_2 z 1 2 + z 2 2 = z 1 z 2
