Further complex numbers — Question 264

ANSWER

(a) A circle centred at $(1,0)$ with radius 2, and the horizontal line $y=1$. (b) $z=1+\sqrt3+i$ or $z=1-\sqrt3+i$. (c) $p(z)=z^2-(2+2i)z-3+2i$.

Worked solution

(a) Let $z=x+iy$. The equations are $(x-1)^2+y^2=4$ and $y=1$. Both are complete loci, with solid boundaries.

(b) Substitute $y=1$ into the circle: $(x-1)^2=3$, so $x=1\pm\sqrt3$.

(c) The root sum is $2+2i$ and product is $(1+i)^2-3=-3+2i$. Thus $p(z)=z^2-(2+2i)z-3+2i$. Its coefficients are complex; conjugate-root pairing is not applicable.

Equivalent justified methods accepted; respect any requested proof method.

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

Part a: Draw the correct labelled circle.

[1 mark]

Part a: Draw the horizontal line $y=1$.

[1 mark]

Part b: Substitute the line into the circle.

[1 mark]

Part b: Find and express both intersections in complex form.

[1 mark]

Part c: Find the root sum and product.

[1 mark]

Part c: Write the correctly expanded quadratic.

[1 mark]

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