Square roots followed by a quadratic shift · source reference

Question

(i) The roots of $w^2=4i$ are $w_1,w_2$. Find $w_1,w_2$ in Cartesian form $x+iy$, showing your working. [3]

(ii) Hence, or otherwise, find in exact Cartesian form the roots $v_1,v_2$ of $v^2-10v+(25-i)=0$. [3]

[2024/RVHS/Prelim/I/5b]

Source working

Printed working

(i)

$$\begin{aligned}(x+iy)^2&=4i\\x^2+2xyi-y^2&=4i\\x^2-y^2+2xyi&=0+4i.\end{aligned}$$

Comparing real and imaginary parts: $x^2-y^2=0\Rightarrow y=\pm x$, and $2xy=4\Rightarrow y=2/x$.

If $y=-x$, $-x=2/x\Rightarrow-x^2=2$. No solutions as $x^2\geq0$ for real $x$.

If $y=x$, $x=2/x\Rightarrow x=\pm\sqrt2\Rightarrow y=\pm\sqrt2$.

The roots are $\sqrt2+\sqrt2i$ and $-\sqrt2-\sqrt2i$.

(ii) Method 1:

$$\begin{aligned}v&=\frac{10\pm\sqrt{(-10)^2-4(25-i)}}2=\frac{10\pm\sqrt{4i}}2\\&=\frac{10\pm(\sqrt2+\sqrt2i)}2\\&=5+\frac{\sqrt2}2+\frac{\sqrt2}2i\quad\text{or}\quad5-\frac{\sqrt2}2-\frac{\sqrt2}2i.\end{aligned}$$

Method 2:

$$\begin{aligned}v^2-10v+(25-i)&=0\\v^2-10v+25&=i\\(v-5)^2&=i\\(2v-10)^2&=4i\\2v-10&=\sqrt2+\sqrt2i\quad\text{or}\quad-\sqrt2-\sqrt2i.\end{aligned}$$

Dividing by 2 and adding 5 gives the same two roots.

Printed answer key

(i) $\sqrt2+\sqrt2i$ or $-\sqrt2-\sqrt2i$. (ii) $5\pm\sqrt2/2\pm(\sqrt2/2)i$ with matching signs.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 1; solution pages 4. These are original document page numbers, retained as text metadata.