Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

If ((1-i)/(1+i))^100 = a+ib, then (1) a = 2, b = -1 (2) a = 1, b = 0 (3) a = 0, b = 1 (4) a = -1, b = 2

Solution:

((1-i)/(1+i)) = ((1-i)(1-i)/(1+i)(1-i))

= (1-2i-1)/(1+1)

= -2i/2

= -i

((1-i)/(1+i))100 = -i100

= (i4)25

= 1

Comparing with a+ib, a = 1, b = 0

Hence option (2) is the answer.

Was this answer helpful?

 
   

0 (0)

(6)
(1)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

Ask
Question