Use the Binomial Theorem to calculate $z^4$ where $z$ is the complex number $2 − i$. Simplify your answer using the fact that $i^2 = −1$. Your final answer should be in the form $a + bi$ where $a$ and $b$ are integers.
I'm confused which version of the binomial theorem to use? What is my $x$ and what is my $y$ in the summation equation?
I am trying to use $(x+y)^n = \displaystyle \sum_{r=0}^n \binom{n}{r} x^{n-r} y^r$
Best Answer
This boils down to treating $x = 2, y = -i.$ We can see that $$(2+(-i))^4 = \sum_{r=0}^4 \binom{4}{r}2^{4-r}(-i)^r$$ $$= \binom{4}{0}2^4(-i)^0 + \binom{4}{1}2^3(-i)^1 + \binom{4}{2}2^2(-i)^2 + \binom{4}{3}2^1(-i)^3 + \binom{4}{4}2^0(-i)^4$$ $$=16 + 4 \times 8 \times (-i) + 6 \times 4 \times (-1) + 4 \times 2 \times i + 1 \times 1 \times 1$$ $$ =16 - 32i - 24 + 8i + 1 = -7-24i.$$