Among the statements (S1) : The set z C--i:|z|=1 and (z-i)/(z+i) is purely real contains e…

Question

Among the statements

(S1) : The set ${z\in C-{-i}:|z|=1$ and $\frac{z-i}{z+i}$ is purely real $}$ contains exactly two elements, and

(S2) : The set ${z\in C-{-1}:|z|=1$ and $\frac{z-1}{z+1}$ is purely imaginary $}$ contains infinitely many elements.

Accepted Answer

$\begin{gathered} \frac{z-i}{z+i}=\frac{\vec{z}+i}{\vec{z}-i} \ =z\vec{z}-i\vec{z}-iz-1=z\vec{z}+zi+i\vec{z}-1 \ =z+\vec{z}=0 \ =2x=0 \ =x=0\, \end{gathered}$

$\begin{gathered} |z|=1 \ 	herefore \, z=i\, (z
e -i) \end{gathered}$

Statement 1 is incorrect

$\begin{gathered} \frac{z-i}{z+i}+\frac{\vec{z}-1}{\vec{z}+1}=0 \ =z\vec{z}-\vec{z}+z-1+z\vec{z}-z+\vec{z}-1=0 \ =z\vec{z}=1 \ =|z|=1 \end{gathered}$

Statement 2 is correct

Solutions · 1