Question

Determine for which value (s) of λ the real part of z = 1+λi/1−λi equals zero

Answer 1

$z=\dfrac{1+\lambda {i}}{1-\lambda{i}}\\ z=\dfrac{(1+\lambda{i})(1+\lambda{i})}{(1-\lambda{i})(1+\lambda{i})}\\ z=\dfrac{1+2\lambda{i}-\lambda^2}{1+\lambda^2}\\ z=\dfrac{1-\lambda^2}{1+\lambda^2}+\dfrac{2\lambda{i}}{1+\lambda^2}\\ \reals(z)=\dfrac{1-\lambda^2}{1+\lambda^2}=0\\ \therefore 1-\lambda^2=0\\ \lambda^2=1\\ \lambda=\pm \sqrt{1}\\ \lambda=\pm1$