Notes: `(1 + i)^4` Use the Binomial Theorem to expand the complex number, then simplify the result.

Tuesday, January 27, 2009

$\displaystyle{{\left({1}+{i}\right)}}^{{4}}$ Use the Binomial Theorem to expand the complex number, then simplify the result.

We are required to expand a binomial that contains a complex number. In order to do this we shall use the binomial theorem. This is used as follows:

where:

a = first term

b = last term

n = exponent (power raised to of the binomial)

k = term number - 1

So let's use the above equation to determine the sum of our binomial:

$\displaystyle{{\left({1}+{i}\right)}}^{{4}}={\sum_{{{k}={1}}}^{{4}}}{\left(\frac{{{4}!}}{{{\left({4}-{k}\right)}!\cdot{k}!}}\right)}\cdot{{a}}^{{{4}-{k}}}\cdot{{b}}^{{k}}$

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

$\displaystyle{{\left({1}+{i}\right)}}^{{4}}={1}+{4}{i}+{6}{{i}}^{{2}}+{4}{{i}}^{{3}}+{{i}}^{{4}}$

Let's revise our complex rules:

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

$\displaystyle{{\left({1}+{i}\right)}}^{{4}}={1}+{4}{i}+{6}{\left(-{1}\right)}+{4}{\left(-{i}\right)}+{1}=-{4}$

ANSWER: -4

Notes

Tuesday, January 27, 2009

$\displaystyle{{\left({1}+{i}\right)}}^{{4}}$ Use the Binomial Theorem to expand the complex number, then simplify the result.

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