Question

Z1/z2 is purely imaginary number then |2z1+3z2/2z1-3z2|

Answer 1

Answer:

   1

Step-by-step explanation:

Step-by-step explanation:

Z₁/Z₂   is purely imaginary number

=> Z₁/Z₂  = ik

(2Z₁ + 3Z₂)/(2Z₁ - 3Z₂)

Lets Divide numerator & Denominator by Z₂

= (2Z₁/Z₂ + 3) /(2Z₁/Z₂ - 3)

= (2k i + 3)/(2k i - 3)

=  (3 + i 2k)/ (-3  + i 2k)

| (3 + i 2k)| = | (3 - i 2k)|  = √9 + 4k²

=> | (3 + i 2k)/ (-3  + i 2k)  | = √9 + 4k² / √9 + 4k²  = 1

=> | (2Z₁ + 3Z₂)/(2Z₁ - 3Z₂) | = 1