Statement 1: (z+i/z-1) is purely real and |z| = 1, then there are exactly 2 complex numbers z.

Question

Consider two statements:

Statement 1: \(\left(\frac{z+i}{z-i}\right)\) is purely real and |z| = 1, then there are exactly 2 complex numbers z.

Statement 2: \(\left(\frac{z+1}{z-1}\right)\) is purely imaginary, then there are infinite such complex numbers z.

Then

(1) Statement 1 is true

(2) Statement 2 is true

(3) Both statement 1 and statement 2 are true

(4) Both statement 1 and statement 2 are false

Answer 1

Correct option is: (3) Both statement 1 and statement 2 are true

Statement 1: \(\left(\frac{-i-z}{i-z}\right)\)

Using rotation

\(\arg \left(\frac{-i-z}{i-z}\right)\) is 0 or \(\pi\)  

\(\Rightarrow (\mathrm{z})\) lies on the circle

\(\Rightarrow\) again infinite such complex numbers.

\(z=\cos \theta+i \sin \theta,\) satisfies.