1.

If `|z^2=1|=|z|^2+1`, then show that `z`lies on the imaginary axis.

Answer» Let z = (x + iy). Then, `z^(2) = (x^(2)-y^(2)) and |z|^(2) = (x^(2) + y^(2))`.
Now, `|z^(2)-1|= |z|^(2) + 1`
`rArr" "|(x^(2)-y^(2)-1)+i(2xy)|=(x^(2)+y^(2)+1)`
`rArr" "|(x^(2)-y^(2)-1)+i(2xy)|^(2)=(x^(2)+y^(2)+1)^(2)`
`rArr" "(x^(2)-y^(2)-1)^(2)+4x^(2)y^(2)=(x^(2)+y^(2)+1)^(2)`
`rArr" "[x^(2)+(y^(2)+1)]^(2)-[x^(2)-(y^(2)+1)]^(2)=4x^(2)y^(2)`
`rArr" "4x^(2)(y^(2)+1)=4x^(2)y^(2) rArr 4x^(2){(y^(2)+1)-y^(2)}=0`
`rArr" "4x^(2)= 0 rArr x = 0`
Hence, z = 0 + iy, which shows that z is imaginary.


Discussion

No Comment Found

Related InterviewSolutions