Answer to Question #254757 in Algebra for inga

Question

it is given that z+2i=iz+$\lambda$, w/z=2+2i, Imw=8, where z and w are complex numbers and $\lambda$ is a real constant. what is the value of $\lambda$

Answer 1

$z+2i=iz+\lambda\\\frac{w}{z}=2+2i\\z(1-i)=\lambda-2i\\z=\frac{\lambda-2i}{1-i}$

Rationalise the denominator

$z=\frac{(\lambda-2i)(1+i)}{(1)^2-(-1)^2}=\frac{\lambda+\lambda i-2i-2i^2}{2}$

$z=\frac{\lambda+2}{2}+\frac{i(\lambda-2)}{2}$

$z=\frac{\lambda+2}{2}+\frac{i(\lambda-2)}{2}$

Given, imaginary part of w=8,

$w=(2+2i)z$

$=(2+2i)\frac{(\lambda+2)+i(\lambda-2)}{2}$

$=(1+i)[(\lambda+2)+i(\lambda-2)]$

$=\lambda+2+i(\lambda-2)+i(\lambda+2)-1(\lambda-2)\\=\lambda+2-\lambda+2+i(\lambda-2+\lambda+2)\\=4+i(2\lambda)$

Since, lm(w)=8

$2\lambda=8\\\lambda=4$