The number of roots of the equation `z^2 = 2 barz` is

Options

A. 2
B. 3
C. 4
D. zero

Answer

C

Explanation

`z^(2)=2barz`
Let `z+x+iyimpliesz^(2)=(x+iy)^(2)=x^(2)-y^(2)+2xyi`
`impliesx^(12)-y^(2)=2x,2xy=-2yimplies2(x+1)y=0.`
`x=-1and y=0 `
`implies(-1)^92)-y^(2)=1(-1)`
`implies1-y^(2)=-2`
`impliesy^(2)=3impliesy=+-sqrt3`
`therefore` Roots are `-1+sqrt3iand -1-sqrt3i.`
and for `y=0, x^(2)-0=2x`
`x(x-2)=0`
`x=0 and 2.`
Hence, roots are `0,2`