Question

Write (1 + i)7 − (1 − i)7 in the form a + bi for some real numbers a and b.

Answer 1

Remeber, the expansion of
`(x+y)^n=\sum_(k=0)^n\binom{n}{k}x^(n-k)y^k`

Then:

`(1+i)^7=\sum_(k=0)^7\binom{7}{k}1^(7-k)i^k=\sum_(k=0)^7\binom{7}{k}i^k\\=\binom{7}{0}i^0+\binom{7}{1}i^1+\binom{7}{2}i^2+\binom{7}{3}i^3+\binom{7}{4}i^4+\binom{7}{5}i^5+\binom{7}{6}i^6+\binom{7}{7}i^7\\=1+7(i)+21(-1)+35(-i)+35(1)+21(i)+7(-1)+1(-i)\\=8-8i`

and

`(1-i)^7=\sum_(k=0)^7\binom{7}{k}1^(7-k)(-i)^k=\sum_(k=0)^7\binom{7}{k}(-1)^ki^k\\=\binom{7}{0}i^0-\binom{7}{1}i^1+\binom{7}{2}i^2-\binom{7}{3}i^3+\binom{7}{4}i^4-\binom{7}{5}i^5+\binom{7}{6}i^6-\binom{7}{7}i^7\\=1-7(i)+21(-1)-35(-i)+35(1)-21(i)+7(-1)-1(-i)\\=8+8i`

Then

`(1+i)^7-(1-i)^7=8-8i-(8+8i)=8-8i-8-8i=0-16i`, where a=0 and b=-16