Answer to Question #218690 in Linear Algebra for Tshego

Question

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

Answer 1

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

If $\lambda$ is a real number, then $\text{Re} \ z=\frac{1-\lambda^2}{1+\lambda^2}=0$

$1-\lambda^2=0$

$\lambda^2=1$

$\lambda=\pm 1$

Answer: $\lambda =\pm1.$