Calculate (1+i)^200=?

Because the exponent 200 could be written as 2*100, we'll
calculate first (1+i)^2.

We'll expand the binomial using
the formula:

(a+b)^2 = a^2 + 2ab +
b^2

a = 1 and b = i

(1+i)^2 =
1^2 + 2*1*i + i^2, where i^2 = -1

(1+i)^2 = 1 + 2i -
1

We'll eliminate like
terms:

(1+i)^2 = 2i

Now, we'll
write (1+i)^200 = [(1+i)^2]^100

 [(1+i)^2]^100 =
(2i)^100

(2i)^100 =
2^100*i^100

We'll recall the power of
i:

i^2 = -1

i^3 = i^2*i = -1*i
= -i

i^4 = i^2*i^2 = (-1)*(-1) =
1

We'll write 100 as a multiple of
4:

100 = 4*25

i^100 =
(i^4)^25

i^4 = 1

i^100 =1^25 =
1

(1+i)^200 = 2^100 = (2^4)^25 =
16^25