2i^275 + (1+i)^-2. write the expression in the form a+ib

Question

Its a complex analysis maths question

Answer 1

2i^275 + (1+i)^-2. write the expression in the form a+ib

there are only 4 values of i:

i = sqrt(-1)

i^2 = -1

i^3 = -i = -sqrt(-1)

i^4 = 1

2i^275 + (1+i)^-2  lets look at the 2i^275 first.  divide 275/4 = 68.75

or 68 3/4  this is the exponent.  (i^4)^68 = 1^64 = 1 with the remainder of i^3

or -i

this becomes 2 (-i) or -2i.

replacing in the expression:  -2i + (1 + i)^-2

-2i + 1/(1 + i)^2  get a common denominator of (1 + i)^2

[-2i(1 + i)^2 + 1]/(1 + i)^2

[-2i(1 + 2i + i^2) + 1] /(1 + 2i + i^2)

[-2i(1 + 2i -1) + 1]/(1 + 2i -1)

[-2i(2i) + 1]/2i

-4i^2 + 1/ 2i