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

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Answer 1 · 2016-03-16T04:09:08

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

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

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

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

$=-i + 5 + 10i - 10 - 5i + 1$

$=-4+4i$

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