Problem source

\begin{questionparts}
\item In the Argand diagram, the points $Q$ and $A$ represent the complex numbers $4+6i$ and $10+2i$. If $A$, $B$, $C$, $D$, $E$, $F$ are the vertices, taken in clockwise order, of a regular hexagon (regular six-sided polygon) with centre $Q$, find the complex number which represents $B$.
\item Let $a$, $b$ and $c$ be real numbers. Find a condition of the form $Aa+Bb+Cc=0$,
where $A$, $B$ and $C$ are integers, which ensures that \[\frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i}\]
is real. 
\end{questionparts}

Solution source

\begin{questionparts}
\item 
\begin{center}
    \begin{tikzpicture}[scale=0.4]
        \draw[->] (-12,0) -- (12,0) node[right] {Re};
        \draw[->] (0,-12) -- (0,12) node[above] {Im};

        \coordinate (A) at (10,2);
        \coordinate (Q) at (4,6);
        \coordinate (B) at ({7+2*sqrt(3)},{4+3*sqrt(3)});
        

        \filldraw (Q) circle (2pt) node[right, above] {$Q = 4+6i$};
        \filldraw (A) circle (2pt) node[right, above] {$A = 10+2i$};
        \draw (B) circle (2pt) node[right, above] {$B$};
        \draw[dashed] (A) -- (Q) -- (B);

        \pic [draw, angle radius=1.2cm, angle eccentricity=1.25, "$\frac{\pi}{3}$"] {angle = A--Q--B};
        
    \end{tikzpicture}
\end{center}

We are looking for $(10+2i) - (4+6i) = 6 - 4i$ rotated by $\frac{\pi}{3}$ and then added to $4+6i$, which is

\begin{align*}
(6-4i)(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) &= (6-4i)\left(\tfrac12 +\tfrac{\sqrt{3}}2i\right) \\
&= 3+2\sqrt{3} + (3\sqrt{3}-2)i
\end{align*}

\item

\begin{align*}
&&& \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} &\in \mathbb{R} \\
\Longleftrightarrow && \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} &= \frac{a}{1-i}+\frac{b}{1-2i}+\frac{c}{1-3i} \\
&& 0 &= a\left ( \frac{1}{1+i} - \frac{1}{1-i} \right)+ b\left ( \frac{1}{1+2i} - \frac{1}{1-2i} \right)+ c\left ( \frac{1}{1+3i} - \frac{1}{1-3i} \right) \\
&&&= a\left ( \frac{(1-i)-(1+i)}{1^2+1^2} \right) + 
b\left ( \frac{(1-2i)-(1+2i)}{1^2+2^2} \right) + c\left ( \frac{(1-3i)-(1+3i)}{1^2+3^2} \right) \\
&&&= -\frac{2i}{2}a-\frac{4i}{5}b-\frac{-6i}{10}c \\
\Longleftrightarrow && 0 &= a+\tfrac45b+\tfrac35c
\end{align*}

\end{questionparts}