Question

How do you write the expression `(1 - i)^5` in the standard form a + bi?

Answer 1

`(1-i)^5 = -4+4i`

Explanation:

The exponent in this case isn't too difficult to calculate directly using the binomial theorem. Proceeding with that, we have:

`(1-i)^5 = sum_(n=0)^5 ((5),(n))1^(n)(-i)^(5-n)=sum_(n=0)^5 ((5),(n))(-i)^(5-n)`

`=(-i)^5 + 5(-i)^(4)+10(-i)^3+10(-i)^2+5(-i)^1+(-i)^0`

`=-i + 5 + 10i - 10 - 5i + 1`

`=-4+4i`