The vertices of a square are `z_1,z_2,z_3 and z_4` taken in the anticlockwise order, then `z_3=`

Options

A. `-iz_(1)+(1+i)z_(2)`
B. `iz_(1)+(1-i)z_(2)`
C. `z_(1)+(1+i)z_(2)`
D. `(1+i)z_(1)+z_(2)`

Answer

A

Explanation

Let ABCD be the given square such that the affixes of the vertices A,B,C and D are `z_(1),z_(2),z_(3)` and `z_(4)` respectively.
The diagonals AC and BD bisect each other.

`therefore (z_(1)+z_(3))/(2) =(z_(2)+z_(4))/(2) rArr z_(1)+z_(3)=z_(2)+z_(4)` .................(i)
Clearly, `bar(AD)` can be obtained by rotating `bar(AB)` through `pi//2`.
`therefore bar(AD)=bar(AB)e^(ipi//2)`
`rArr z_(4)-z_(1)=(z_(2)-z_(1))e^(ipi//2)`
`rArr z_(4)-z_(1) = (z_(2)-z_(1))i rArr z_(4)=iz_(2)+(1-i)z_(1)`
Putting `z_(4)=iz-(2)+(1-i)z_(1)` in (i), we get `z_(3)=-iz_(1)+(1-i)z_(2)`.