Question

If `z_1, z_2 in C ,z1 2 in R ,z_1(z1 2-3z2 2)=2a n dz_2(3z1 2-z2 2)=11` , then the value of `z1 2+z2 2` is `10` b. `12` c. `5` d. `8`

Answer 1

`z_1(z_1^2 + 3z_2^2) = 2` (1)
`z_2(3z_1^2 - z_2^2) = 11` (2)
multiplying eqn (2) with i and adding (1) and (20
`z_1^3- 3z_1z_2^2 + i(3z_1^2z_2 - z_2^3)= 2+ 11i`
now, `(z_1+ iz_2)^3= 2+ 11i` (A)
now we will`(1) - i(2):`
`z_1^3 - 3z_1z_2^2 - i(3z_1^2z_2 - z_2^3) = 2- 11i`
`(z_1-iz_2)^3 = 2-11i` (B)
multiplying eqn (A) & (B)
`(z_1+iz_2)^3 (z_1-iz_2)^3 = (2+11i)(2-11i)`
`[(z_1+iz_2)(z_1-iz_2)]^3 = (2)^2 - (i11)^2`
`= [z_1^2 - (iz_2)^2]^3 = 4-i^2(121)`
`=(z_1^2 - i^2z_2^2)^3 = 4+121= 125`
`= z_1^2 +z_2^2 = (125)^(1/3) = 5`
option C is correct