Euler's Identity - HW help - Thank you!

kingcumin22 said:
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To do the one must know that:
\(\displaystyle \begin{gathered}
{e^x} = \sum\limits_{k = 0}^\infty {\frac{{{x^k}}}{{k!}}} \hfill \\
\sin (x) = \sum\limits_{k = 0}^\infty {\frac{{{x^{2k + 1}}}}{{(2k + 1)!}}} \hfill \\
\cos (x) = \sum\limits_{k = 0}^\infty {\frac{{{x^{2k}}}}{{(2k)!}}} \hfill \end{gathered} \)
Note that the \(\displaystyle \sin(x)\) has all odd powers while the \(\displaystyle \cos(x)\) has all even powers.

Now look at the powers of \(\displaystyle e^x\)
\(\displaystyle \begin{array}{*{20}{l}} {{i^1} = i}&{{i^2} = - 1}&{{i^3} = - i}&{{i^4} = 1} \\
{{i^5} = i}&{{i^6} = - 1}&{{i^7} = - i}&{{i^8} = 1} \\
{{i^9} = i}&{{i^{10}} = - 1}&{{i^{11}} = i}&{{i^{12}} = 1} \\
{{i^{13}} = i}&{{i^{14}} = - 1}&{{i^{15}} = - i}&{{i^{16}} = 1} \end{array}\)
Note the cyclic nature of the powers. They repeat in a four-cycle.
All the even powers are either \(\displaystyle \pm 1\) and all the odd powers are either \(\displaystyle \pm i\).
This enables us to evaluate powers of \(\displaystyle i\) say \(\displaystyle i^{55}\) Well divide \(\displaystyle {55}\) by four and the remainder is \(\displaystyle 3\).
Thus \(\displaystyle i^{55}=i^3=-i\).
These are preliminary facts to help you answer the question.
Please post your work.
 
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