If `arg(z^(3//8))=(1)/(2)arg(z^(2)+barz^(1//2))`, then which of the following is not possible ?

Options

A. `|z|=1`
B. `z=barz`
C. `arg(z)=0`
D. None of these

Answer

D

Explanation

`(d)` We have
`arg(z^(3//8))=(1)/(2)arg(z^(2)+barzz^(1//2))`
`implies2arg(z^(3//8))=arg(z^(2)+barzz^(1//2))`
`impliesarg(z^(3//4))-arg(z^(2)+barzz^(1//2))=0[:'2arg(z)=arg(z^(2))]`
`impliesarg((z^(2)+barzz^(1//2))/(z^(3//4)))=0`
`impliesIm(z^(5//4)+(barz)/(z^(1//4)))=0`
`impliesz^(5//4)+(barz)/(z^(1//4))=bar((z^(5//4)+(barz)/(z^(1//4))))`
`impliesz^(5//4)+(barz)/(z^(1//4))=(barz)^(5//4)+(z)/((barz)^(1//4))`
`impliesz^(5//4)+(barz(barz)^(1//4))/(|z|^(1//2))=(barz)^(5//4)+(zz^(1//4))/(|z|^(1//2))`
`impliesz^(5//4)-(barz)^(5//4)=(z^(5//4)-(barz)^(5//4))/(|z|^(1//2))`
`implies{z^(5//4)-(barz)^(5//4)}(1-(1)/(|z|^(1//2)))=0`
`:. z=barz` or `|z|=1`.