Q22 of 153 Page 13

If |z + 1| = z + 2(1 + i), find z.

Given:


⇒ |z+1|=z+2(1+i)


Let us assume z=x+iy


⇒ |x+iy+1|=x+iy+2+2i


⇒


Equating Real and Imaginary parts on both sides


⇒ y+2=0


⇒ y=-2----------------(1)


⇒


⇒ (x+1)2+y2=(x+2)2


⇒ x2+2x+1+(-2)2=x2+4x+4


⇒ 2x=1+4-4


⇒ 2x=1


⇒ .


∴ .


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