Find the value of m if `((1 + i)/(1 - i))^m = 1`

Answer

a,c,

Explanation

`because (x^(2) + 1 + (1)/(x^(2)))^(n) = ((1 + x^(2) + x^(4))^(n))/(x^(2n))`
` = (a_(0) + a_(1)x^(2) + a_(2)x^(4) + ...+ a_(2n) x^(4n))/(x^(2n))`
` therefore ` Number of terms = 2n+ 1
Term indepandent of ` x = a_(n)` = Constant term in ` (x^(2) + 1 + (1)/(x^(2)))^(n)`
= Coefficient of ` x^(2n) " in " (1 + x+ x^(2))^(n)`
` (d^(n))/(dx^(n)) (1 + x + x^(2))^(n) ne 2^(n-1)`
Coefficient of ` x^(2n-2) " in" (x^(2) + 1 + x (1)/(x^(2)))^(n)`
= Coefficient of ` x^(4n -2) " in " (1 + x ^(2) + x^(4) )^(n)`
= Coefficient of ` x^(2n-1) " in " (1 + x + x^(2))^(n)`
Now ,let ` (1 + x + x^(2))^(n) = lambda _(0) + lambda_(1)x + lambda _(2) x^(2) + ...+ lambda_(2n -1) x^(2lambda -1) + lambda _(2n) x^(2n)`
On replacing x by ` (1)/(x)` , we get
`(1 + (1)/(x) + (1)/(x^(2)))^(n) = lambda_(0) + (lambda_(1))/(x) + (lambda_(2))/(x^(2)) + ...+ (lambda _(2n -1))/(x^(2n -1) + )(lambda _(2n))/(x^(2n))`
or ` (1 + x + x^(2))^(n) = lambda_(2n) + lambda_(2n-1) x + ...+ lambda_(1) x^(2n-1) + lambda_(0) x^(2n)`
On differntiating both sides w.r.t.x we get
`n (1 + x + x^(2))^(n). (1 + 2x ) = lambda_(2n-1) + ...+ 2n lambda_(0) x^(2n-1)`
On putting x = 0 , we get ` lambda_(2n-1) = n `
Hence , coeffcient of ` x^(2n-2) = n `
and coefficient of `x^(2) " in " (x^(2) + 1 + (1)/(x^(2)))^(n)`
= Coefficient of ` x^(2n+2) " in" (1 + x^(2) + x^(4))^(n)`
= Coefficient of ` x^(n+1) " in " (1 + x + x^(2))^(n)`
` = (d^(n+1))/(dx^(n+1)) (1 + x + x^(2))^(n) ne n `