VITEEE 2010 Paper

Given : |z| - z =1+2i
If z = x + iy, then this equation reduces to
| x + iy | - (x + iy) = 1 + 2i
⇒ (√x2+y2−x)+(−iy)=1+2i
On comparing real and imaginary parts of both sides of this equation, we get
√x2+y2−x=1
⇒ √x2+y2=1+x
⇒x2+y2=(1+x)2
⇒x2+y2=1+x2+2x
⇒y2=1+2x...(i)
and -y = 2 ⇒ y = -2
Putting this value in eq.(i), we get
(−2)2=1+2x
⇒ 2x=3⇒x=

3

2

∴‌z=x+iy=

3

2

−2i