Question

The complex number satisfying the equation $\overline{z}-(1+i)z-(3+2i)\overline{z}+(1+5i)=0$ are

• A. $1+i$
• B. $\frac{1}{2}+\frac{i}{2}$
• C. $\frac{3}{2}-i\frac{7}{6}$
• D. $3-2i$

Answer 1

The correct option is C $\frac{3}{2}-i\frac{7}{6}$

let $z=x+iy$ then

$x-iy+(1+i)(x+iy)-(3+2i)(x-iy)+(1+5i)=0$

$⟹x-iy+x-y+iy+ix-3x+3yi-2xi-2y+1+5i=0$

$⟹(x+x-3x-2x+1-y-2y)+(-y+y+x+3y-2x+5)i=0$

Hence,

$3x+3y-1=0$......[1]

$x-3y-5=0$........[2]

Solving [1] and [2], we get:

$x=3/2$ and $y=-7/6$

Hence $z=\frac{3}{2}-i\frac{7}{6}$