A modulus equation imposes a sign · source reference

Question

The complex numbers $z,w$ satisfy $2z+1=|w|$ and $2w-z=4+24i$. Find $z,w$, giving answers in the form $a+ib$, where $a,b$ are real numbers. [4]

[2023/CJC/Prelim/I/1]

Source working

Printed working — three methods

Method 1: $2z+1=|w|$ (1), $2w-z=4+24i$ (2). From (2), $z=2w-4-24i$. Substitute into (1): $2(2w-4-24i)+1=|w|$, so $4w-7-48i=|w|$.

Let $w=a+bi$:

$$4(a+bi)-8-48i+1=\sqrt{a^2+b^2},$$ $$(4a-7)+(4b-48)i=\sqrt{a^2+b^2}.$$

Imaginary parts: $4b-48=0\Rightarrow b=12$. Real parts:

$$4a-7=\sqrt{a^2+b^2}=\sqrt{a^2+12^2}\quad(**).$$ $$(4a-7)^2=a^2+144\Rightarrow15a^2-56a-95=0.$$

Hence $a=-19/15$ or $5$. From (**), $4a-7$ is positive. At $a=-19/15$, $4(-19/15)-7<0$; reject this value. Therefore $a=5,b=12,z=2(5+12i)-4-24i=6$, and $w=5+12i$.

Method 2: $2z+1=|w|$ is a positive real number. Put $z=x,w=a+bi$. Equation (2) gives $2a-x=4$ and $2b=24\Rightarrow b=12$. Equation (1) gives $2x+1=\sqrt{a^2+b^2}$ (3). Substituting $x=2a-4,b=12$:

$$2(2a-4)+1=\sqrt{a^2+12^2},$$ $$4a-7=\sqrt{a^2+12^2},\quad(4a-7)^2=a^2+144,$$ $$16a^2-56a+49=a^2+144\Rightarrow15a^2-56a-95=0.$$

Thus $a=-19/15$ or $5$, and $x=-98/15$ or $6$. Since $2z+1$ is positive, $x=-98/15$ fails: $2(-98/15)+1<0$. Reject $x=-98/15,a=-19/15$. Therefore $x=6,a=5,b=12$.

Method 3 as printed: From (2), $2w=4+z+24i$. The printed working asserts $z$ is positive real because $|w|$ is positive, then obtains

$$2|w|=\sqrt{(4+z)^2+24^2}\quad(3).$$

$(1)-(3)$ is printed as $4z+2=\sqrt{(4+z)^2+24^2}$. Squaring:

$$(4z+2)^2=(4+z)^2+24^2,$$ $$16z^2+16z+4=16+8z+z^2+576,$$ $$15z^2+8z-588=0,$$ $$z=\frac{-8\pm\sqrt{8^2+4(15)(588)}}{2(15)}=6\quad\text{or}\quad-98/15.$$

Since $z>0$, the source chooses $z=6$, then $w=5+12i$.

Qualification of printed method 3

The inference $|w|>0\Rightarrow z>0$ is insufficient: (1) initially gives $z>-1/2$, not $z>0$. That weaker valid condition still rejects $-98/15$ and preserves the answer. The source’s “$(1)-(3)$” label is not subtraction; the equation is formed by substituting $2|w|=4z+2$.

Printed answer key

$z=6,w=5+12i$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 1; solution pages 5, 6. These are original document page numbers, retained as text metadata.