InterviewSolution
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InterviewSolution
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If `z and w` are two complex numbers simultaneously satisfying te equations, `z^3+w^5=0 and z^2 +overlinew^4 = 1,` thenA. `z` and `w` both are purely realB. `z` is purely real and `w` is purely imagineryC. `w` is purely real and `z` is purely imagineryD. `z` and `w` both are imaginery |
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Answer» Correct Answer - A `(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` |
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