Complex Numbers
12.0 Rotation
12.0 Rotation
If ${z_1}$, ${z_2}$ and ${z_3}$ are the three vertices of a triangle $ABC$ taken in counter-clockwise direction, then
$$\begin{equation} \begin{aligned} \frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}} = \frac{{OQ}}{{OP}}\left( {\cos \alpha + i\sin \alpha } \right) \\ {\text{ = }}\frac{{CA}}{{BA}}.{e^{i\alpha }} = \frac{{\left| {{z_3} - {z_1}} \right|}}{{\left| {{z_2} - {z_1}} \right|}}.{e^{i\alpha }} \\\end{aligned} \end{equation} $$
and $$\arg \left( {\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}}} \right) = \alpha $$
where $\alpha $ is the angle through which $OP/AB$ is rotated in anti-clockwise direction so that it coincides with $OQ/CA$.
Question 12. If $A(2+3i)$ and $B(3+4i)$ are two vertices of a square $ABCD$, taken in anti-clockwise order, then find $C$ and $D$.
Solution: Let us assume the affix of $C$ and $D$ be ${z_3}$ and ${z_4}$ respectively. As shown in figure, $\angle DAB = {90^ \circ }$ and $AD=AB$.
Apply rotation for points $B$, $A$ and $D$, we get $$\begin{equation} \begin{aligned} \frac{{{z_4} - (2 + i3)}}{{(3 + i3) - (2 + i3)}} = \frac{{AD}}{{AB}}{e^{i\frac{\pi }{2}}} \\ \Rightarrow {z_4} - (2 + i3) = (1 + i)i \\ \Rightarrow {z_4} = 2 + 3i + i - 1 = 1 + 4i \\\end{aligned} \end{equation} $$
Now, apply rotation for points $A$, $B$ and $C$, we get $$\begin{equation} \begin{aligned} \frac{{{z_3} - (3 + i4)}}{{(2 + i3) - (3 + i4)}} = \frac{{CB}}{{AB}}{e^{ - i\frac{\pi }{2}}} \\ \Rightarrow {z_3} = 3 + i4 - (1 + i)( - i) \\ \Rightarrow {z_3} = 2 + i5 \\\end{aligned} \end{equation} $$
