complex no.:

solve the equation:

mod z i.eIzI=z+1+2i

Assume Z=x+iy

Mod(Z)=root(x^2+y^2)

root(x^2+y^2)=x+iy+1+2i

root(x^2+y^2)=(x+1)+i(y+2)

equating real and imaginary parts

y+2=0

y=-2

root(x^2+y^2)=x+1

x=3/2

ANS=Z=3/2-2i

IzI=z+1+2i

let z=x+iy

therfore, (x²+y²)^1/2=x+iy+1+2i

                               =(x+1)+(y+2)i

               since LHS is purely real, therefore

y+2=0, ie.  y=0

also, (x²+y²)^1/2=x+1

squaring both the sides and putting y=-2

=>  x²+4=x²+2x+1

=>   x=3/2

therfore the answer is x=3/2, y=-2.