Complex Numbers
    8.0 De Moivre's Theorem

8.0 De Moivre's Theorem
Statement $1$: If $n$ is any integer, then

(i) ${(\cos \theta + i\sin \theta )^n} = \cos (n\theta ) + i\sin (n\theta )$

(ii) $(\cos {\theta _1} + i\sin {\theta _1})(\cos {\theta _2} + i\sin {\theta _2})(\cos {\theta _3} + i\sin {\theta _3})...(cos{\theta _n} + i\sin {\theta _n}) = \cos ({\theta _1} + {\theta _2} + {\theta _3} + ...{\theta _n}) + i\sin ({\theta _1} + {\theta _2} + {\theta _3} + ...{\theta _n})$


Statement $2$: If $p,q \in Z$ and $q \ne 0$ then
$${(\cos \theta + i\sin \theta )^{\frac{p}{q}}} = \cos \left( {\frac{{2k\pi + p\theta }}{q}} \right) + i\sin \left( {\frac{{2k\pi + p\theta }}{q}} \right)$$
where $k = 0,1,2,3...,q - 1$


Question 9. Simplify $$\frac{{{{\left[ {2(\cos \theta - i\sin \theta )} \right]}^{10}}{{\left( {\cos \theta + i\sin \theta } \right)}^{20}}}}{{{{\left( {\cos \theta - i\sin \theta } \right)}^{18}}}}$$

Solution: $$\begin{equation} \begin{aligned} \Rightarrow \frac{{{2^{10}}{{\left[ {\cos \left( { - \theta } \right) + i\sin \left( { - \theta } \right)} \right]}^{10}}{{\left( {\cos \theta + i\sin \theta } \right)}^{20}}}}{{{{\left[ {\cos ( - \theta ) + i\sin ( - \theta )} \right]}^{18}}}} \\ \Rightarrow \frac{{{2^{10}}\left[ {\cos \left( { - 10\theta } \right) + i\sin \left( { - 10\theta } \right)} \right]\left( {\cos 20\theta + i\sin 20\theta } \right)}}{{\left[ {\cos ( - 18\theta ) + i\sin ( - 18\theta )} \right]}} \\ \Rightarrow \frac{{{2^{10}}\left[ {\cos \left( {20\theta - 10\theta } \right) + i\sin \left( {20\theta - 10\theta } \right)} \right]}}{{\left[ {\cos ( - 18\theta ) + i\sin ( - 18\theta )} \right]}} \\ \Rightarrow \frac{{{2^{10}}\left[ {\cos \left( {10\theta } \right) + i\sin \left( {10\theta } \right)} \right]}}{{\left[ {\cos ( - 18\theta ) + i\sin ( - 18\theta )} \right]}} \\ \Rightarrow {2^{10}}\cos \left( {10\theta - ( - 18\theta )} \right) + i\sin \left( {10\theta - ( - 18\theta )} \right) \\ \Rightarrow {2^{10}}\cos \left( {28\theta } \right) + i\sin \left( {28\theta } \right) \\\end{aligned} \end{equation} $$
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