Let `A(2,0)` and `B(z)` are two points on the circle `|z|=2`. `M(z')` is the point on `AB`. If the point `barz'` lies on the median of the triangle `OAB` where `O` is origin, then `arg(z')` is

Options

A. `tan^(-1)((sqrt(15))/(5))`
B. `tan^(-1)(sqrt(15))`
C. `tan^(-1)((5)/(sqrt(15)))`
D. `(pi)/(2)`

Answer

A

Explanation

`(a)` `M(z')` is mid point of `AB`, so `z'=(z+2)/(2)`

`impliesbarz'=(barz+2)/(2)`
Now `z`, `1`,`(barz)/(2)+1` are collinear
`implies|{:(z,barz,1),((barz)/(2)+1,(z)/(2)+1,1),(1,1,1):}|=0`
`impliesz((z)/(2)+1-1)-barz((barz)/(2)+1-1)+1((barz)/(2)-(z)/(2))=0`
`implies(z^(2))/(2)-(barz)^(2)/(2)+(barz-z)/(2)=0`
`implies(z-barz)(z+barz-1)=0`
`impliesz-barz=0` or `(z+barz-1)=0`
`impliesz+barz=1` or `Re(z)=(1)/(2)`
`|z|=2implies(1)/(4)+(Im(z))^(2)=4`
`impliesIm(z)=(sqrt(15))/(2)`
`impliesz=(1)/(2)+(isqrt(15))/(2)`
`impliesarg(z')=tan^(-1)((sqrt(15))/(5))`