Notes: `(5 - sqrt(3) i )^4` Use the Binomial Theorem to expand the complex number, then simplify the result.

Thursday, March 18, 2010

$\displaystyle{{\left({5}-\sqrt{{{3}}}{i}\right)}}^{{4}}$ Use the Binomial Theorem to expand the complex number, then simplify the result.

You need to use the binomial formula, such that:

$\displaystyle{{\left({x}+{y}\right)}}^{{n}}={\sum_{{{k}={0}}}^{{n}}}{\left(\matrix{{n}\\{k}}\right)}{{x}}^{{{n}-{k}}}{{y}}^{{k}}$

You need to replace 5 for x, $\displaystyle\sqrt{{3}}\cdot{i}$ for y and 4 for n, such that:

$\displaystyle{{\left({5}-\sqrt{{3}}\cdot{i}\right)}}^{{4}}={4}{C}{0}{{\left({5}\right)}}^{{4}}+{4}{C}{1}{{\left({5}\right)}}^{{3}}\cdot{{\left(-\sqrt{{3}}\cdot{i}\right)}}^{{1}}+{4}{C}{2}{{\left({5}\right)}}^{{2}}\cdot{{\left(-\sqrt{{3}}\cdot{i}\right)}}^{{2}}+{4}{C}{3}{{\left({5}\right)}}^{{1}}\cdot{{\left(-\sqrt{{3}}\cdot{i}\right)}}^{{3}}+{4}{C}{4}{3}{a}\cdot{{\left(-\sqrt{{3}}\cdot{i}\right)}}^{{4}}$

By definition, nC0 = nCn = 1, hence $\displaystyle{4}{C}{0}={4}{C}{4}={1}.$

By definition nC1 = nC(n-1) = n, hence $\displaystyle{4}{C}{1}={4}{C}{3}={4}.$

By definition $\displaystyle{n}{C}{2}=\frac{{{n}{\left({n}-{1}\right)}}}{{2}},$ hence $\displaystyle{4}{C}{2}={6}.$

$\displaystyle{{\left({5}-\sqrt{{3}}\cdot{i}\right)}}^{{4}}={625}-{500}\sqrt{{3}}\cdot{i}+{450}\cdot{{i}}^{{2}}-{60}\sqrt{{3}}\cdot{{i}}^{{3}}+{9}\cdot{{i}}^{{4}}$

Using the powers of i yields:

$\displaystyle{i}={i};{{i}}^{{2}}=-{1},{{i}}^{{3}}=-{i},{{i}}^{{4}}={1}$

$\displaystyle{{\left({5}-\sqrt{{3}}\cdot{i}\right)}}^{{4}}={184}-{440}\sqrt{{3}}\cdot{i}$

Hence, expanding the complex number using binomial theorem yields the simplified result $\displaystyle{{\left({5}-\sqrt{{3}}\cdot{i}\right)}}^{{4}}={184}-{440}\sqrt{{3}}\cdot{i}$

Notes

Thursday, March 18, 2010

$\displaystyle{{\left({5}-\sqrt{{{3}}}{i}\right)}}^{{4}}$ Use the Binomial Theorem to expand the complex number, then simplify the result.

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