Considering $ (1+i)^n - (1 - i)^n $, Complex Analysis I have been working on problems from Complex Analysis by Ahlfors, and I got stuck in the following problem:
Evaluate:
$$
(1 + i)^n - (1-i)^n 
$$
I have just "reduced" to:
$$ 
(1 + i)^n - (1-i)^n  = \sum_{k=0} ^n i^k(1 - (-1)^k)
$$
by using expansion of each term.
Thanks.

ok. we can use the binomial theorem. we will check out for small values of $n.$
case $n = 1, \ z_1 = 1+ i -(1-i) = 2i$
case $n = 2, \ z_2 = (1+i)^2 - (1-i)^2 =(1 + 2i + i^2) -(1 - 2i + i^2) = 2(2i) = 4i $
case $n = 3, \ z_2 = (1+i)^3 - (1-i)^3 =(1 + 3i + 3i^2 + i^3) -(1 - 3i + 3i^2 - i^3) = 2(3i +i^3) = 4i$
case $n = 4, \ z_4 = (1 + i)^4 - (1 - i)^4 =2(4i + 4i^3) = 0$
case $n = 5, \ 
z_5 = (1 + i)^5 - (1 - i)^5 =2(5i + 10i^3 + i^5) = -8i $
case $n = 6, \ 
z_6 = (1 + i)^6 - (1 - i)^6 =2(6i + 20i^3 + 6i^5) = -16i$
case $n = 7, \ 
z_7 = (1 + i)^7 - (1 - i)^7 =2(7i + 35i^3 + 21i^5 + i^7) = -16i$
case $n = 8, \ 
z_8 = (1 + i)^8 - (1 - i)^8 =2(8i + 56i^3 + 56i^5 + 8i^7) = 0$
so it looks like the formula can be put in four classes: 
if $n$ is a multiple of $4,$ then $z_n = 0$ 
if $n = 4k+1,$ then $z_n = (2i)^{(n+1)/2}$
if $n = 4k+2, 4k+3,$ then $z_n = 2(2i)^{(n+1)/2}$
