Use the Maclaurin series to prove that $e^{i\theta} = cos(\theta) + isin(\theta)$

Question

Our teacher assigned extra credit for which we are allowed to use any source (including this website) to find the answer. This material has not been covered in class or in the book we are using. We learned Taylor series using series of real numbers but we have not covered anything remotely similar to the problem below so I have no idea how to even begin this problem.

Let i =$\sqrt{−1}$
a) Use the Taylor series centered at x=0 to prove that, for all $\theta$,
$e^{i\theta} = cos(\theta) + isin(\theta)$

b) Use above answer to prove that $[cos(\theta)+ isin(\theta)]^n = cos(n\theta)+ isin(n\theta)]$

c) We may write $[cos(\theta)+ isin(\theta)]^n = cos^n(x)(1+itan(\theta))^n$
Use the 5th order Taylor polynomial for $(1+x)^\frac{1}{3}$ with $x = itan(\theta)$ to find an estimate for $sin\left(\frac{\theta}{3}\right)$.
Use this approximation to estimate the value of $sin\left(\frac{\pi}{18}\right)$.
Hint: Two complex numbers $a+bi$ and $c+di$ are equal if $a=c$ and $b=d$.

Answer 1

Take the series expansion of $e^z$, which is defined also for complex $z$ (and actually it is used to define the principal value of $e^z$, which is the usual acception for writing $e^z$). Split it into the even and odd components. $$ e^{\,z} = \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }} {{k!}}} = \sum\limits_{0\, \leqslant \,k} {\left( {\frac{{z^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}} + \frac{{z^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} \right)} = \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} $$ Then putting $iy$ in place of $z$, you will notice that $$ \begin{gathered} e^{\,i\,y} = \sum\limits_{0\, \leqslant \,k} {\frac{{i^{\,\left( {2k} \right)} y^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{i^{\,\left( {2k + 1} \right)} y^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} = \hfill \\ = \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - 1} \right)^{\,k} y^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - 1} \right)^{\,k} i\,y^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} = \hfill \\ = \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - 1} \right)^{\,k} y^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}}} + i\sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - 1} \right)^{\,k} \,y^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} = \hfill \\ = \cos y + i\sin y \hfill \\ \end{gathered} $$

Answer 2

Your question is a bit more than others asked so it is not quite a duplicate. For the first part, look at $e^{i\theta}$ $=$ $\cos \theta + i \sin \theta$, a definition or theorem? .

For (b), use (a): you have $e^{i \theta}=\cos \theta + i \; \sin \theta$. Then $(\cos \theta+i \; \sin \theta)^n=(e^{i \theta})^n$. Use properties of exponents and then think about your 'new' exponent like you would have in part (a).

For (c), use $(e^{i \theta})^n=(\cos \theta + i \sin \theta)^n$ and try factoring out a $\cos$ from within the parenthesis and see what you might get. For the last part of it, you already have a good hint. Just follow what you already did in (c) in 'reverse' until you get to the $e^{i \theta}$ part and see what you get (you may want to reference (a) again and relate real and imaginary parts).