Question

If 2z1/3z2 is a purely imaginery number then z1-z2/z1+z2 is

Answer 1

Answer:

give proper qustion plasea

Answer 2

Answer:

Since, 2z1/3z2 = k'i ...(Given)

or, z1/z2= (3k'/2) i

or, z1/z2= ki (where 3k'/2=k) ...(1)

|(z1-z2)/ (z1+z2)| = |z1{1-(z2/z1)}/z1{1+(z2/z1)}|

Cancelling z1's, above equation becomes

|(z1-z2)/ (z1+z2)| = |[1-(z2/z1)]/[1+(z2/z1)]|

= | (1-ki)/ (1+ki) | ...(using (1))

= |(1-ki)|/ |(1+ki)|

= 1