Find the imaginary part of the complex number(1+i)^10 + (1-i)^10.

We'll recall the rectangular form of a complex
number:

z = a + b*i, where a is the real part and b is the
imaginary part.

We notice that we can write (1+i)^10 =
[(1+i)^2]^5

We'll expand the
binomial:

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

We'll replace i^2 by
-1:

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

We'll
eliminate like terms:

(1+i)^2 =
2i

(1+i)^10 = (2i)^5 =
2^5*i^5

We know that i^5 = i^4*i = 1*i =
i

(1+i)^10 = 32 i (1)

We
notice that we can write (1-i)^10 = [(1-i)^2]^5

We'll
expand the binomial:

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

(1-i)^2 = -2i

(1-i)^10 =
(-2i)^5 = -32i (2)

We'll add (1) +
(2):

(1-i)^2 + (1-i)^10 = 32 i - 32
i

(1-i)^2 + (1-i)^10 = 0

We
can write the result as a complex number, as it
follows:

(1-i)^2 + (1-i)^10 = 0 +
0i

We notice that the imaginary part of the
complex number is b = 0.