If `z and w` are two complex numbers simultaneously satisfying te equations, `z^3+w^5=0 and z^2 +overlinew^4 = 1,` then

Options

A. `z` and `w` both are purely real
B. `z` is purely real and `w` is purely imaginery
C. `w` is purely real and `z` is purely imaginery
D. `z` and `w` both are imaginery

Answer

A

Explanation

`(a)` `z^(3)=-omega^(5)implies|z|^(3)=|omega|^(5)implies|z|^(6)=|omega|^(10)` ………`(i)`
and `z^(2)=(1)/(baromega^(4))implies|z|^(2)=(1)/(|omega|^(4))implies|z|^(6)=(1)/(|omega|^(12))`……..`(ii)`
From `(i)` and `(ii)`, `|omega|=1` and `|z|=1implieszbarz=omegabaromega=1`
Again `z^(6)=omega^(10)`
and `z^(6)*baromega^(12)=1`
`z^(6)=(1)/(baromega^(12))=omega^(10)` (using `(iii)`)
`implies (omegabaromega)^(10)(baromega)^(2)=1`
`implies(baromega)^(2)=1`
`implies baromega=1` or `-1impliesomega=1` or `-1`
If `omega=1`, then `z^(3)+1=0` and `z^(2)=1impliesz=-1`
If `omega=-1`, then `z^(3)-1=0` and `z^(2)=1impliesz=1`
Hence, `z=1` and `omega=-1` or `z=-1` and `omega=1`