Answer to Question #252556 in Algebra for moe

Question

It is given that 

$z+2i=iz+\lambda,\quad\quad$ $\frac{w}{z}=2+2i,\quad\quad$ $\mathrm{Im}w=8,$

where z and w are complex numbers, and λ is a real constant. Which of the following is the value of \lambdaλ?

λ=4

λ=8

λ=3

λ=−3

λ=−4

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, ln(w)=8

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