Problem source

Prove that
\[
(\cos\theta +\mathrm{i}\sin\theta)
(\cos\phi +\mathrm{i}\sin\phi)
=
\cos(\theta+\phi) +\mathrm{i}\sin(\theta+\phi)
\]
and that, for every positive integer $n$,
$$
{(\cos {\theta} + \mathrm{i}\sin {\theta})}^n
= \cos{n{\theta}} + \mathrm{i}\sin{n{\theta}}.
$$

By considering 
$(5-\mathrm{i})^2(1+\mathrm{i})$,
or otherwise,
prove that
\[
\arctan\left(\frac{7}{17}\right)+2\arctan\left(\frac{1}{5}\right)=\frac{\pi}{4}\,.
\]
Prove also that 
\[
3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{1}{20}\right)+\arctan\left(\frac{1}{1985}\right)=\frac{\pi}{4}\,.
\]
[Note that $\arctan\theta$ is another notation for $\tan^{-1}\theta$.]

Solution source

\begin{align*}
&& LHS &= (\cos\theta +\mathrm{i}\sin\theta)
(\cos\phi +\mathrm{i}\sin\phi) \\
&&&= \cos \theta \cos \phi - \sin \theta \sin \phi + \mathrm{i}(\sin \theta \cos \phi + \cos \theta \sin \phi) \\
&&&= \cos (\theta + \phi) + \mathrm{i} \sin (\theta + \phi) \\
&&&= RHS
\end{align*}

Therefore we can see $(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n \theta$.

\begin{align*}
&& (5-i)^2(1+i) &= (24-10i)(1+i) \\
&&&= (24+10) + i(24-10) \\
&&&= 34+14i \\
\Rightarrow && 2\arg(5-i) +\arg(1+i) &= \arg(34+14i) \\
\Rightarrow && 2\arctan\left (-\frac{1}{5} \right) + \frac{\pi}{4} &= \arctan \left ( \frac{7}{17} \right) \\
\Rightarrow && 2\arctan\left (\frac{1}{5} \right) +\arctan \left ( \frac{7}{17} \right) &= \frac{\pi}{4} \\
\end{align*}

Consider $(1+i)(4-i)^3(20-i)$

\begin{align*}
&& (1+i)(4-i)^3(20-i) &= (21+19i)(52-47i) \\
&&&= 1985+i \\
\Rightarrow && \frac{\pi}{4} - 3 \arctan \left ( \frac{1}{4} \right) -\arctan \left ( \frac{1}{20} \right) &= \arctan \left ( \frac{1}{1985} \right)
\end{align*}