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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 $(x+y)^n = \displaystyle \sum_{r=0}^n \binom{n}{r} x^{n-r} y^r$

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  • $\begingroup$ Please state which version(s) of the binomial theorem you know. $\endgroup$ – David Feb 18 '16 at 2:19
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    $\begingroup$ $(a+b)^4=a^4+\binom{4}{1}a^3b+\binom{4}{2}a^2b^2+\binom{4}{3}ab^3+b^4$. $\endgroup$ – André Nicolas Feb 18 '16 at 2:24
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    $\begingroup$ Let $x = 2$; let $y = -i$. $\endgroup$ – N. F. Taussig Feb 18 '16 at 2:25
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This boils down to treating $x = 2, y = -i.$ We can see that $$(2+(-i))^4 = \sum_{r=0}^4 \binom{4}{r}2^{4-r}(-i)^r$$ $$= \binom{4}{0}2^4(-i)^0 + \binom{4}{1}2^3(-i)^1 + \binom{4}{2}2^2(-i)^2 + \binom{4}{3}2^1(-i)^3 + \binom{4}{4}2^0(-i)^4$$ $$=16 + 4 \times 8 \times (-i) + 6 \times 4 \times (-1) + 4 \times 2 \times i + 1 \times 1 \times 1$$ $$ =16 - 32i - 24 + 8i + 1 = -7-24i.$$

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