Root midpoint and a parallelogram · source reference

Question

The complex number $z$ satisfies $z^2+(2-4i)z=c$, where $c$ is a complex constant.

(i) Given $z_1,z_2$ are two distinct roots, show $(z_1+z_2)/2=-1+2i$. [2]

It is given $c=4i$ and $\operatorname{Im}(z_1)<\operatorname{Im}(z_2)$.

(ii) Find $z_1,z_2$ in the form $x+iy$, with $x,y$ real. Leave $x,y$ exact. [3]

(iii) On an Argand diagram indicate clearly $P_1,P_2$ representing $z_1,z_2$ respectively. [2]

(iv) O and Q represent 0 and $w$ respectively. $OP_1P_2Q$ is a parallelogram. Using (iii), find $w$. [2]

[2022/VJC/Promo/9]

Source working

Printed working

(i) $(z-z_1)(z-z_2)=z^2+(2-4i)z-c$. Comparing the coefficient of $z$, $-z_1-z_2=2-4i$, hence $(z_1+z_2)/2=-1+2i$.

Alternative quadratic-formula proof:

$$z_1=\frac{-(2-4i)-\sqrt{(2-4i)^2-4(-c)}}2\quad(1),$$ $$z_2=\frac{-(2-4i)+\sqrt{(2-4i)^2-4(-c)}}2\quad(2).$$

Adding the two expressions cancels the radicals:

$$z_1+z_2=\frac{-2(2-4i)}2,\qquad\frac{z_1+z_2}2=\frac{-(2-4i)}2=-1+2i.$$

The printed label says “(1)−(2)” above this addition; this is a label typo, not the operation used.

(ii)

$$\begin{aligned}z&=\frac{-(2-4i)\pm\sqrt{(2-4i)^2-4(-4i)}}2\\&=\frac{-2(1-2i)\pm\sqrt{-12-16i+16i}}2\\&=\frac{-2(1-2i)\pm2\sqrt{-3}}2\\&=-(1-2i)\pm\sqrt3i=-1+(2\pm\sqrt3)i.\end{aligned}$$

Thus $z_1=-1+(2-\sqrt3)i$, $z_2=-1+(2+\sqrt3)i$.

(iii) The native diagram labels O, Re(z), Im(z), $P_1,P_2,Q$, $-1$, $2-\sqrt3$, $2$, $2+\sqrt3$ and $2\sqrt3$. Its two marked half-segments between $P_1$ and $P_2$ show the midpoint at height 2.

(iv) $P_1P_2=OQ$ and $P_1P_2=2+\sqrt3-(2-\sqrt3)=2\sqrt3$. Hence $w=2\sqrt3i$.

Teacher note

Q lies on the imaginary axis since $OP_1P_2Q$ is a parallelogram.

Printed answer key

(ii) $z_1=-1+(2-\sqrt3)i,z_2=-1+(2+\sqrt3)i$. (iv) $w=2\sqrt3i$.

Native diagrams

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 2; solution pages 7, 8. These are original document page numbers, retained as text metadata.