What is `((1+i)^(4n+5))/((1-i)^(4n+3))` equal to, where n is a natural number and `i=sqrt(-1)` ?

{1+(-i)^(4n+3)}(1-i), (ninN) equals

((1+i)/(1-i))^(4n+1) where n is a positive integer.

For positive integers n_(1), n_(2) the value of the expression (1+i)^(n_(1)) + (1+ i^(3))^(n_(1)) + (1+ i^(5))^(n_(2)) + (1+ i^(7))^(n_(2)) , where i= sqrt-1 is a real number if and only if

What is the modulus of the complex number i^(2n+1) (-i)^(2n-1) , where n in N and i=sqrt(-1) ?

If (4+sqrt(15))^n=I+f, where n is an odd natural number, I is an integer and ,then

For positive integers n_1,n_2 , the value of the expression : (1+i)^(n_1)+(1+i^3)^(n_1) +(1+i^5)^(n_2)+ (1+i^7)^(n_2) , where i= sqrt(-1) is a real number if and only if :

Show that (-sqrt(-1))^(4n+3) =i , where n is a positive integer.

Show that (-sqrt(-1))^(4n+3)=i , where n is a positive integer.

If a and b are positive integers such that N=(a+ib)^(3)-107i (where N is a natural number), then the value of a is equal to (where i^(2)=-1 )