Home / IBDP Maths analysis and approaches Topic: AHL 1.12 Complex numbers HL Paper 1

# IBDP Maths analysis and approaches Topic: AHL 1.12 Complex numbers HL Paper 1

### Question:

Consider the complex numbers z1 = 1 + bi and z2 = (1 – b2) – 2bi, where b ∈ R, b ≠ 0.

(a) Find an expression for z1z2 in terms of b.

Ans: z1z2  = (1 + bi) ((1-b2) – (2b)i)

= (1-b2 – 2i2b2) + i (-2b + b – b3)

= (1 + b2) + i (-b – b3)

Note: Award A1 for 1+ b2 and A1 for − bi – b3i .

(b) Hence, given that arg (z1 z2) = $$\frac{\pi }{4}$$ , find the value of b.

Ans: arg (z1z2) = arctan $$\left ( \frac{-b-b^{3}}{1 + b^{2}} \right )= \frac{\pi }{4}$$

EITHER

arctan (-b) = $$\frac{\pi }{4}$$ (since 1+b2 ≠, 0 for b ∈ R)

OR

-b – b3 = 1 + b2  (or equivalent)

THEN

b =−1

### Question:

Consider the complex numbers z1 = 1 + bi and z2 = (1 – b2) – 2bi, where b ∈ R, b ≠ 0.

(a) Find an expression for z1z2 in terms of b.

Ans: z1z2  = (1 + bi) ((1-b2) – (2b)i)

= (1-b2 – 2i2b2) + i (-2b + b – b3)

= (1 + b2) + i (-b – b3)

Note: Award A1 for 1+ b2 and A1 for − bi – b3i .

(b) Hence, given that arg (z1 z2) = $$\frac{\pi }{4}$$ , find the value of b.

Ans: arg (z1z2) = arctan $$\left ( \frac{-b-b^{3}}{1 + b^{2}} \right )= \frac{\pi }{4}$$

EITHER

arctan (-b) = $$\frac{\pi }{4}$$ (since 1+b2 ≠, 0 for b ∈ R)

OR

-b – b3 = 1 + b2  (or equivalent)

THEN

b =−1

### Question

Consider the complex numbers Z1 = cos $$\frac{11\pi }{12}+sin\frac{11\pi }{12}$$ and Z2 = cos $$\frac{\pi }{6}+isin\frac{\pi }{6}$$

(a)  (i) Find $$\frac{Z_1}{Z_2}$$

Ans: $$\frac{z_{1}}{z_{2}}= cos(\frac{11\pi }{12}-\frac{\pi }{6})+i sin (\frac{11\pi }{12}-\frac{\pi }{6}) = cos \frac{3\pi }{4}+i sin \frac{3\pi }{4}$$

(ii) Find $$\frac{Z_2}{Z_1}$$ [3]

Ans: $$\frac{z_{2}}{z_{1}} = cos\frac{3\pi }{4}-isin \frac{3\pi }{4}$$

(b) $$\frac{Z_1}{Z_2}$$ and $$\frac{Z_2}{Z_1}$$ bare represented by three points O, A and B respectively on an Argand diagram. Determine the area of the triangle OAB. [2]

Ans: valid attempt to calculate area of their triangle (angle between OA and OB is $$\frac{\pi }{2}\Rightarrow area (=\frac{1}{2}\times 1\times 1)=\frac{1}{2}$$

### Question

If $${z_1} = a + a\sqrt 3 i$$ and $${z_2} = 1 – i$$, where a is a real constant, express $${z_1}$$ and $${z_2}$$ in the form $$r\,{\text{cis}}\,\theta$$, and hence find an expression for $${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6}$$ in terms of a and i.

### Markscheme

$${z_1} = 2a{\text{cis}}\left( {\frac{\pi }{3}} \right){\text{, }}{z_2} = \sqrt 2 {\text{ cis}}\left( { – \frac{\pi }{4}} \right)$$     M1     A1     A1

EITHER

$${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = \frac{{{2^6}{a^6}{\text{cis(0)}}}}{{{{\sqrt 2 }^6}{\text{cis}}\left( {\frac{\pi }{2}} \right)}}\left( { = 8{a^6}{\text{cis}}\left( { – \frac{\pi }{2}} \right)} \right)$$     M1     A1     A1

OR

$${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = {\left( {\frac{{2a}}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{7\pi }}{{12}}} \right)} \right)^6}$$     M1     A1

$$= 8{a^6}{\text{cis}}\left( { – \frac{\pi }{2}} \right)$$     A1

THEN

$$= – 8{a^6}{\text{i}}$$     A1

Note: Accept equivalent angles, in radians or degrees.

Accept alternate answers without cis e.g. $${\text{ = }}\frac{{8{a^6}}}{{\text{i}}}$$

[7 marks]

### Question

Given that z is the complex number $$x + {\text{i}}y$$ and that $$\left| {\,z\,} \right| + z = 6 – 2{\text{i}}$$ , find the value of x

and the value of y .

### Markscheme

$$\sqrt {{x^2} + {y^2}} + x + y{\text{i}} = 6 – 2{\text{i}}$$     (A1)

equating real and imaginary parts     M1

$$y = – 2$$     A1

$$\sqrt {{x^2} + 4} + x = 6$$     A1

$${x^2} + 4 = {(6 – x)^2}$$     M1

$$– 32 = – 12x \Rightarrow x = \frac{8}{3}$$     A1

[6 marks]

### Question

Given that $$(4 – 5{\text{i}})m + 4n = 16 + 15{\text{i}}$$ , where $${{\text{i}}^2} = – 1$$, find m and n if

a. m and n are real numbers;[3]

Ans: attempt to equate real and imaginary parts     M1

equate real parts: $$4m + 4n = 16$$; equate imaginary parts: $$-5m = 15$$     A1

$$\Rightarrow m = -3,{\text{ }}n = 7$$     A1

[3 marks]

b. m and n are conjugate complex numbers.[4]

let $$m = x + {\text{i}}y,{\text{ }}n = x – {\text{i}}y$$     M1

$$\Rightarrow (4 – 5{\text{i}})(x + {\text{i}}y) + 4(x – {\text{i}}y) = 16 + 15{\text{i}}$$

$$\Rightarrow 4x – 5{\text{i}}x + 4{\text{i}}y + 5y + 4x – 4{\text{i}}y = 16 + 15{\text{i}}$$

attempt to equate real and imaginary parts     M1

$$8x + 5y = 16,{\text{ }} -5x = 15$$     A1

$$\Rightarrow x = -3,{\text{ }}y = 8$$     A1

$$( \Rightarrow m = -3 + 8{\text{i}},{\text{ }}n = -3 – 8{\text{i}})$$

[4 marks]

### Question

Consider the complex numbers

a. $${z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2}$$ and $${z_2} = – 1 + \sqrt 3 {\text{i }}$$ .

(i)     Write down $${z_1}$$ in Cartesian form.

Ans: $${z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2} \Rightarrow {z_1} = – 2\sqrt 3 {\text{i}}$$     A1

(ii)     Hence determine $${({z_1} + {z_2})^ * }$$ in Cartesian form.[3]

Ans: $${z_1} + {z_2} = – 2\sqrt 3 {\text{i}} – 1 + \sqrt 3 {\text{i}} = – 1 – \sqrt 3 {\text{i}}$$     A1

$${({z_1} + {z_2})^ * } = – 1 + \sqrt 3 {\text{i}}$$     A1

[3 marks]

b. (i)     Write $${z_2}$$ in modulus-argument form.

Ans:  $$\left| {{z_2}} \right| = 2$$

$$\tan \theta = – \sqrt 3$$     (M1)
$${z_2}$$ lies on the second quadrant
$$\theta = \arg {z_2} = \frac{{2\pi }}{3}$$
$${z_2} = 2{\text{cis}}\frac{{2\pi }}{3}$$     A1A1
(ii)     Hence solve the equation $${z^3} = {z_2}$$ .[6]

Ans: attempt to use De Moivre’s theorem     M1

$$z = \sqrt[3]{2}\,{\text{cis}}\frac{{\frac{{2\pi }}{3} + 2k\pi }}{3},{\text{ }}k = 0{\text{, 1 and 2}}$$
$$z = \sqrt[3]{2}\,{\text{cis}}\frac{{2\pi }}{9},\,\sqrt[3]{2}\,{\text{cis}}\frac{{8\pi }}{9},\,\sqrt[3]{2}\,{\text{cis}}\frac{{14\pi }}{9}\left( { = \sqrt[3]{2}\,{\text{cis}}\left( {\frac{{ – 4\pi }}{9}} \right)} \right)$$     A1A1
Note: Award A1 for modulus, A1 for arguments.
Note: Allow equivalent forms for z .
[6 marks]
c. Let $$z = r\,{\text{cis}}\theta$$ , where $$r \in {\mathbb{R}^ + }$$  and $$0 \leqslant \theta < 2\pi$$ . Find all possible values of and $$\theta$$ ,

(i)     if $${z^2} = {(1 + {z_2})^2}$$;

Ans: METHOD 1

$${z^2} = {\left( {1 – 1 + \sqrt 3 {\text{i}}} \right)^2} = – 3\left( { \Rightarrow z = \pm \sqrt 3 {\text{i}}} \right)$$     M1

$$z = \sqrt 3 \,{\text{cis}}\frac{\pi }{2}{\text{ or }}{z_1} = \sqrt 3 \,{\text{cis}}\frac{{3\pi }}{2}\left( { = \sqrt 3 \,{\text{cis}}\left( {\frac{{ – \pi }}{2}} \right)} \right)$$     A1A1

so $$r = \sqrt 3 {\text{ and }}\theta = \frac{\pi }{2}{\text{ or }}\theta = \frac{{3\pi }}{2}\left( { = \frac{{ – \pi }}{2}} \right)$$

Note: Accept $$r\,{\text{cis}}(\theta )$$ form.

METHOD 2

$${z^2} = {\left( {1 – 1 + \sqrt 3 {\text{i}}} \right)^2} = – 3 \Rightarrow {z^2} = 3{\text{cis}}\left( {(2n + 1)\pi } \right)$$     M1

$${r^2} = 3 \Rightarrow r = \sqrt 3$$     A1

$$2\theta = (2n + 1)\pi \Rightarrow \theta = \frac{\pi }{2}{\text{ or }}\theta = \frac{{3\pi }}{2}{\text{ (as }}0 \leqslant \theta < 2\pi )$$     A1

Note: Accept $$r\,{\text{cis}}(\theta )$$ form.

(ii)     if $$z = – \frac{1}{{{z_2}}}$$.[6]

Ans: METHOD 1

$$z = – \frac{1}{{2{\text{cis}}\frac{{2\pi }}{3}}} \Rightarrow z = \frac{{{\text{cis}}\pi }}{{2{\text{cis}}\frac{{2\pi }}{3}}}$$     M1

$$\Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}$$

so $$r = \frac{1}{2}{\text{ and }}\theta = \frac{\pi }{3}$$     A1A1

METHOD 2

$${z_1} = – \frac{1}{{ – 1 + \sqrt 3 {\text{i}}}} \Rightarrow {z_1} = – \frac{{ – 1 – \sqrt 3 {\text{i}}}}{{\left( { – 1 + \sqrt 3 {\text{i}}} \right)\left( { – 1 – \sqrt 3 {\text{i}}} \right)}}$$     M1

$$z = \frac{{1 + \sqrt 3 {\text{i}}}}{4} \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}$$

so $$r = \frac{1}{2}{\text{ and }}\theta = \frac{\pi }{3}$$     A1A1

[6 marks]

d. Find the smallest positive value of n for which $${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} \in {\mathbb{R}^ + }$$ .[4]
Ans: $$\frac{{{z_1}}}{{{z_2}}} = \sqrt 3 \,{\text{cis}}\frac{{5\pi }}{6}$$     (A1)

$${\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} = {\sqrt 3 ^n}{\text{cis}}\frac{{5n\pi }}{6}$$     A1

equating imaginary part to zero and attempting to solve     M1

obtain n = 12     A1

Note: Working which only includes the argument is valid.

[4 marks]

### Question

(z + 2i) is a factor of 2z3 – 3z2 + 8z – 12. Find the other two factors.

If $$(z + 2i)$$ is a factor then $$(z – 2i)$$ is also a factor. $$(z+2i)(z-2i)=(z^{2}+4)$$