Problem source

\begin{questionparts}
	\item If $z=x+\mathrm{i}y,$ with $x,y$ real,
	show that 
	\[
	\left|x\right|\cos\alpha+\left|y\right|\sin\alpha\leqslant\left|z\right|\leqslant\left|x\right|+\left|y\right|
	\]
	for all real $\alpha.$
	\item By considering $(5-\mathrm{i})^{4}(1+\mathrm{i}),$ show that
	\[
	\frac{\pi}{4}=4\tan^{-1}\left(\frac{1}{5}\right)-\tan^{-1}\left(\frac{1}{239}\right).
	\]
	Prove similarly that 
	\[
	\frac{\pi}{4}=3\tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{1}{20}\right)+\tan^{-1}\left(\frac{1}{1985}\right).
	\]
\end{questionparts}

Solution source

\begin{questionparts}
\item If $z=x+iy$ then $|z|^2 = x^2 + y^2 \leq x^2 + y^2 + 2|x||y| \leq (|x|+|y|)^2$.

The LHS is Cauchy-Schwarz with the vectors $\begin{pmatrix} |x| \\ |y| \end{pmatrix}$ and $\begin{pmatrix} \cos \alpha \\ \sin \alpha \end{pmatrix}$, although that's not in the spirit of the question. 

Consider $e^{i \alpha}z = (\cos \alpha x - \sin \alpha y) + i(\sin \alpha x + \cos \alpha y)$ then $\left | \textrm{Re}(e^{i \alpha} z) \right | \leq |z|$ for all values of $\alpha$ and in particular we can choose $\alpha$ to match the signs of the $x$ and $y$ to prove the result in question.

\item Consider $(5-\mathrm{i})^{4}(1+\mathrm{i})$, then 
\begin{align*}
\arg \l (5-\mathrm{i})^{4}(1+\mathrm{i}) \r &= \arg (5-i)^4 + \arg (1+i) \\
&= 4 \arg (5-i) + \arg (1+i) \\
&= -4 \tan^{-1} \frac{1}{5} + \tan^{-1} 1 \\
\\
&= \arg ( (24 - 10i)^2 (1+i)) \\
&= \arg (4 (12 - 5i)^2(1+i)) \\
&= \arg ((119 - 120i)(1+i)) \\
&= \arg (239 - i) \\
&= -\tan^{-1} \frac{1}{239}
\end{align*}

Therefore $\displaystyle \frac{\pi}{4} =4 \tan^{-1} \frac{1}{5}- \tan^{-1} \frac{1}{239}$

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

\begin{align*}
\arg \l(4-i)^3(1+i)(20-i) \r &= -3 \tan^{-1} \frac14 + \tan^{-1} 1 -\tan^{-1} \frac1{20} \\
\\
&= \arg \l(15-8i)(4-i)(1+i)(20-i) \r \\
&= \arg \l (52 - 47i)(1+i)(20-i) \r \\
&= \arg \l (99 + 5i)(20-i) \r \\
&= \arg (1985+i) \\
&= \tan^{-1} \frac1{1985}
\end{align*}

Therefore $\displaystyle \frac{\pi}{4}=3\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{1}{20}+\tan^{-1}\frac{1}{1985}$

\end{questionparts}