z+2i=iz+λzw=2+2iz(1−i)=λ−2iz=1−iλ−2i
Rationalise the denominator
z=(1)2−(−1)2(λ−2i)(1+i)=2λ+λi−2i−2i2
z=2λ+2+2i(λ−2)
z=2λ+2+2i(λ−2)
Given, imaginary part of w=8,
w=(2+2i)z
=(2+2i)2(λ+2)+i(λ−2)
=(1+i)[(λ+2)+i(λ−2)]
=λ+2+i(λ−2)+i(λ+2)−1(λ−2)=λ+2−λ+2+i(λ−2+λ+2)=4+i(2λ)
Since, lm(w)=8
2λ=8λ=4
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