CAIE A-Level Mathematics A2 3.9 Complex Numbers Questions

State $z \omega=(-3+3 \sqrt{3})+(3+3 \sqrt{3}) \mathrm{i}$
Or exact equivalent with real and imaginary parts collected. Need brackets around the coefficient of i.

Marking guidance:

Allow for a=, b= stated correctly.

Obtain $|z|=\sqrt{2}$
Obtain $\arg z=-\frac{\pi}{4}$ final answer
Obtain $|\omega|=6$
Obtain $\arg \omega=\frac{2 \pi}{3}$ final answer

Note: The question does not require the diagram.
If they use $\frac{5 \pi}{12}$ they need to demonstrate where it comes from.
Complex number equivalent to A B is $3 \sqrt{3}+3 \mathrm{i}$.

Show |O A|=|A B|=6, hence isosceles
One mark for 'isosceles' and one mark for 'right angle'.
There will be alternatives e.g. use of Pythagoras (ratio of lengths is

$\angle A O B=\arg \omega-\arg z \omega=-\arg z=\frac{\pi}{4}$ hence third angle is a right
angle
$1: 1: \sqrt{2}$ ), expressing each number in "vector" form and using
scalar product or explaining the effect of multiplying by 1-i.

$\arg z \omega=\arg z+\arg \omega\left(=\frac{2 \pi}{3}-\frac{\pi}{4}=\frac{5 \pi}{12}\right)$
For showing correct use of their angles from part (b).
Must demonstrate where $\frac{5 \pi}{12}$ comes from.

$\arg z \omega=\tan ^{-1} \frac{3+3 \sqrt{3}}{-3+3 \sqrt{3}}$
Correct method for their z $\omega$ from part (a). Must link to point B on diagram or to $\arg z \omega$.

Need to see $\tan ^{-1} \frac{3+3 \sqrt{3}}{-3+3 \sqrt{3}}$ or $\tan \theta=\frac{3+3 \sqrt{3}}{-3+3 \sqrt{3}}$ and not just $\tan ^{-1} \frac{1+\sqrt{3}}{-1+\sqrt{3}}$.

$\Rightarrow \tan \left(\frac{5}{12} \pi\right)=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
Obtain given answer from full and correct working.

Substitute $x=-1+\sqrt{7} \mathrm{i}$ in the equation and attempt expansions of $x^{2}$ and
$x^{3}$

*M1

Use $\mathrm{i}^{2}=-1$ correctly at least once and solve for k
$2(20-4 \sqrt{7} \mathrm{i})+3(-6-2 \sqrt{7} \mathrm{i})+14(-1+\sqrt{7} \mathrm{i})+k=0$

Obtain answer k=-8
SC B1 only for those who show no working for the cube
and square and obtain answer k=-8.

Alternative method for question 10(a)

Attempt division by $(x+1-\sqrt{7} \mathrm{i})$ as far as $2 x^{2}+z_{1} x+\ldots$

*M1

See division on next page.

Use $\mathrm{i}^{2}=-1$ correctly at least once and obtain $2 x^{2}+z_{1} x+z_{2}+$ remainder
Obtain answer k=-8

State answer $-1-\sqrt{7} \mathrm{i}$
Can be seen simply stated on its own, or in a list of roots.

Marking guidance:

Allow if stated clearly in part 10(a).

Carry out a method for finding a quadratic factor with zeros $-1+\sqrt{7} \mathrm{i}$ and
$-1-\sqrt{7} \mathrm{i}$
Or state $(x-(-1+\sqrt{7} \mathrm{i}))(x-(-1-\sqrt{7} \mathrm{i}))(2 x-p)$

Obtain $x^{2}+2 x+8$
Or obtain $(-1+\sqrt{7} \mathrm{i})(-1-\sqrt{7} \mathrm{i}) p=-8$
Or obtain $(-1+\sqrt{7} \mathrm{i})+(-1-\sqrt{7} \mathrm{i})+\frac{p}{2}=-\frac{3}{2}$

Obtain root $x=\frac{1}{2}$, or equivalent, via division or inspection
Needs to follow from the working.

|z-u|=2 is the circle of radius 2 centred at u. Since
$$u=-1+\sqrt7\,i,$$
the locus is a circle centred at $(-1,\sqrt7)$ on the Argand diagram with radius 2.

B1 for centre $-1+\sqrt7 i$. B1 for a circle of radius 2, with scale or radius indicated;
if more than one circle is shown, maximum B1.

Carry out a complete method for calculating the maximum value of $\arg z$ for
correct circle
e.g. $\frac{\pi}{2}+\tan ^{-1} \frac{1}{\sqrt{7}}+\frac{\pi}{4}$ Can be implied by $155.7^{\circ}$.

Obtain answer 2.72 radians
CAO. The question requires radians.