A complex square gives an exact tangent · source reference

Question

$z=\sqrt3+i$, $w=-1+i\sqrt3$. Sketch an Argand diagram with origin O showing Z, W and P representing $z,w,z+w$ respectively. Describe the geometrical shape OWPZ. [3]

By considering the argument of $z+w$, deduce $\tan(5\pi/12)=2+\sqrt3$. [3]

[2010/MJC/Prelim/I/11]

Source working

Printed working

$$z+w=(\sqrt3-1)+i(1+\sqrt3).$$

Addition makes OWPZ a parallelogram.

$$\arg w=\pi-\tan^{-1}(\sqrt3/1)=2\pi/3,$$ $$\angle WOZ=\arg w-\arg z=2\pi/3-\tan^{-1}(1/\sqrt3)=2\pi/3-\pi/6=\pi/2.$$

$OZ=|z|=2$, $OW=|w|=|-1+i\sqrt3|=2$. Since $\angle WOZ=\pi/2$ and $OZ=OW$, OWPZ is a square.

Alternative solution: $w=-1+i\sqrt3=iz=i(\sqrt3+i)$, so $\arg w=\pi/2+\arg z$.

OP represents $z+w$:

$$\arg(z+w)=\arg z+\angle POZ=\pi/6+\pi/4=5\pi/12.$$ $$\begin{aligned}\tan(5\pi/12)&=\frac{\sqrt3+1}{\sqrt3-1}=\frac{(\sqrt3+1)^2}{3-1}\\&=\frac{4+2\sqrt3}{2}=2+\sqrt3.\end{aligned}$$

Alternative angle proof: OP bisects $\angle ZOW$ since OWPZ is a square.

$$\arg(w+z)=\tfrac12(\arg w+\arg z)=\tfrac12(2\pi/3+\pi/6)=5\pi/12.$$

Diagram labels

Both native square diagrams retain O,W,P,Z, Re,Im; ticks $-1,\sqrt3-1,\sqrt3$ horizontally and $1,\sqrt3,1+\sqrt3$ vertically. The second includes OP, the perpendicular projection of P to Re, angle $\pi/4$ between OZ and OP and $5\pi/12$ from Re to OP; P is labelled as representing $z+w$.

Printed answer key

Square.

Native diagrams

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 4; solution pages 20, 21. These are original document page numbers, retained as text metadata.