Further complex numbers — Question 263

ANSWER

(a) $\sqrt3+i$, $-\sqrt3+i$ and $-2i$. (b) The triangle has area $3\sqrt3$ square units.

Worked solution

(a) $8i=8\operatorname{cis}(\pi/2)$. The roots are $2\operatorname{cis}(\pi/6+2k\pi/3)$ for $k=0,1,2$, yielding the three stated Cartesian coordinates.

(b) The horizontal side between $\sqrt3+i$ and $-\sqrt3+i$ has length $2\sqrt3$. Its perpendicular distance from $-2i$ is 3, so $A=\tfrac12(2\sqrt3)(3)=3\sqrt3$. See the solution diagram.

Equivalent justified methods accepted; respect any requested proof method.

Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.

Part a: Find modulus 2.

[1 mark]

Part a: Find three arguments separated by $2\pi/3$.

[1 mark]

Part a: Convert all roots to exact Cartesian form.

[1 mark]

Part b: Plot the three roots accurately and label them.

[1 mark]

Part b: Identify base $2\sqrt3$ and perpendicular height 3.

[1 mark]

Part b: Calculate area $3\sqrt3$.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.