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

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1 Answer

Mar 16, 2016

#(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#

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