Find n

(1) (1-i)ⁿ=iⁿ

(2) (1+i)ⁿ=(1-i)ⁿ

2)(1-i)ⁿ=(1+i)ⁿ

cis(-npi/4)=cis(npi/4)      (2^n/2(1/rt2+i/rt2)ⁿ) 

2isin(npi/4)=0

npi=4m(pi)

n=4m

n should be multiple of 4.

cheeerrrsss

2. (1+i)ⁿ = (1-i)ⁿ

So,

((1+i)/(1-i))ⁿ = 1

((1+i)²/2)ⁿ = 1

therefore,

(2i/2)ⁿ = 1

iⁿ = 1

HENCE, n is a multiple of 4.......!!!!

Sorry i was mistyped the first one ,it is (1-i)ⁿ=2^n.Now please solve it

n =0???

 so  (1- i)ⁿ  =2ⁿ        ==>> [( 1-i)/2]ⁿ    =1    ===>(1/ )ⁿ  [(1 - i)/ root 2]ⁿ  = 1       

    ====>( 1/2)^n/2 cos (-n pi /4)  + i sin ( - pi / 4)    = 1      equATING REAL AND IMAGINARY PARTS

(1/2 )^n/2  cos (n pi/4)   = 1               

(1/2 )n/2 sin  ( n pi /4 ) = 0               ==> sin (n pi )/4  = 0    => n pi /4 =  x  pi          n = 4 x

putting this in (1/2)^n/2  cos (n pi /4)  = 1     ==>   (  1/2 )^2x cos( x pi) = 1     ====> cos ( x pi ) = 2 ^2x    

  but cos x pi  = 1, -1                   2^2x not equal to -1  

                                                   2 ^2x = 1 ==> x =0                ====> n = 4x   = 0

rate if useful     :)

answer can only be n=0

(1-i)⁰=1,2⁰=1

checking:-

(|1-i|)ⁿ=|2|ⁿ

(2^(1/2))ⁿ=2ⁿ

which can only be true for n=0

2. (1+i)ⁿ = (1-i)ⁿ

So,

((1+i)/(1-i))ⁿ = 1

((1+i)²/2)ⁿ = 1

therefore,

(2i/2)ⁿ = 1

please rate if correct

iⁿ = 1

HENCE, n is a multiple of 4.......!!!!

good ques