Question #252556

It is given that 


z+2i=iz+λ,z+2i=iz+\lambda,\quad\quad wz=2+2i,\frac{w}{z}=2+2i,\quad\quad Imw=8,\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


1
Expert's answer
2021-10-19T17:19:02-0400

z+2i=iz+λwz=2+2iz(1i)=λ2iz=λ2i1iz+2i=iz+\lambda\\\frac{w}{z}=2+2i\\z(1-i)=\lambda-2i\\z=\frac{\lambda-2i}{1-i}

Rationalise the denominator

z=(λ2i)(1+i)(1)2(1)2=λ+λi2i2i22z=\frac{(\lambda-2i)(1+i)}{(1)^2-(-1)^2}=\frac{\lambda+\lambda i-2i-2i^2}{2}


z=λ+22+i(λ2)2z=\frac{\lambda+2}{2}+\frac{i(\lambda-2)}{2}


z=λ+22+i(λ2)2z=\frac{\lambda+2}{2}+\frac{i(\lambda-2)}{2}


Given, imaginary part of w=8)


w=(2+2i)zw=(2+2i)z


=(2+2i)(λ+2)+i(λ2)2=(2+2i)\frac{(\lambda+2)+i(\lambda-2)}{2}


=(1+i)[(λ+2)+i(λ2)]=(1+i)[(\lambda+2)+i(\lambda-2)]


=λ+2+i(λ2)+i(λ+2)1(λ2)=λ+2λ+2+i(λ2+λ+2)=4+i(2λ)=\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

2λ=8λ=42\lambda=8\\\lambda=4


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