Question

Previous year IIT jee Question
Chapter :- complex number and quadratic equation.​

Answer 1

Step-by-step explanation:

$\small \leadsto \tt( \frac{1 + i}{1 - i} ) ^{x} = 1 \\ \: \sf Multiply(1 + i) \: on \: the \: given \: two \: fractions \\ \small \leadsto \tt \: [ \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)}] ^{x} = 1 \\ \tt \small \leadsto \: [\frac{1 + 2n + {i}^{2} }{1 - {i}^{2} } ]^{x} = 1 \\ \small \leadsto \: \tt [ \frac{2i}{1 + 1} ] ^{x} = 1 \\ \leadsto \tt \: {i}^{x} = 1$

x=4n for n ∈ I.

So,

x=4n, where n is any positive integer, So,

[Option 'A' is correct]

Answer 2

Answer:

Answer in the attachment.