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Complex Numbers

Practice exam-style IB Math AA questions for Complex Numbers, aligned with the syllabus and grouped by topic.

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Question 1
HL • Paper 1
Easy
Non Calculator
HL • Paper 1
Easy
Non Calculator

Let z=x+yiz=x+yi, where x,y∈Rx,y\in\mathbb{R}, and suppose that (1+i)z=4+2i(1+i)z=4+2i.

A

zz in Cartesian form.

[3]
B

∣z∣|z| and the principal argument of zz.

[2]
Question 2
HL • Paper 1
Easy
Non Calculator
HL • Paper 1
Easy
Non Calculator

Let z=1−iz=1-i.

A

z6z^6 in Cartesian form.

[4]
Question 3
HL • Paper 2
Easy
Calculator Permitted
HL • Paper 2
Easy
Calculator Permitted

The complex number zz is defined by

z=5−3i2+iz=\frac{5-3{\text{i}}}{2+{\text{i}}}
A

Express zz in the form a+bia+b{\text{i}}, where a,b∈Qa,b\in\mathbb{Q}.

[2]
B

the exact value of ∣z∣|z|.

[1]
C

the principal argument of zz, in radians.

[2]
Question 4
HL • Paper 2
Easy
Calculator Permitted
HL • Paper 2
Easy
Calculator Permitted

Let z=cos⁡θ+isin⁡θz=\cos\theta+{\text{i}}\sin\theta, where θ∈R\theta\in\mathbb{R}.

A

Expand z3z^3 in Cartesian form.

[3]
B

cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta.

[1]

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Question 5
HL • Paper 2
Easy
Calculator Permitted
HL • Paper 2
Easy
Calculator Permitted

The polynomial

P(z)=z3−7z2+17z−15P(z)=z^3-7z^2+17z-15

has real coefficients. One root of P(z)=0P(z)=0 is 2+i2+{\text{i}}.

A

another non-real root of P(z)=0P(z)=0.

[1]
B

the remaining root of P(z)=0P(z)=0.

[3]
Question 6
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A complex number z=x+yiz=x+yi, where x,y∈Rx,y\in\mathbb{R}, satisfies
∣z−(2+i)∣=∣z+1−3i∣|z-(2+i)|=|z+1-3i|.

A

the Cartesian equation satisfied by xx and yy.

[3]
B

the locus represented by this equation in the Argand diagram.

[1]
Question 7
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The polynomial P(z)=z3+pz2+qz+10P(z)=z^3+pz^2+qz+10 has real coefficients. One root is 1+2i1+2i and the third root is real.

A

another non-real root of P(z)P(z).

[1]
B

the real root.

[2]
C

the values of pp and qq.

[2]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let z=2cis⁡(π6)z=2\operatorname{cis}\left(\frac{\pi}{6}\right) and w=3−iw=\sqrt{3}-i.

A

Write ww in modulus-argument form.

[2]
B

zw2zw^2 in Cartesian form.

[3]

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Question 9
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let u=1+iu=1+i and v=3+iv=\sqrt{3}+i.

A

uv\frac{u}{v} in Euler form.

[3]
B

geometrically the transformation represented by multiplying a complex number by uv\frac{u}{v}.

[2]
Question 10
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let zz be a non-zero complex number such that ∣z∣=2|z|=\sqrt{2} and 0<arg⁡z<π20<\arg z<\frac{\pi}{2}. It is given that the principal argument of (1+i)z2(1+i)z^2 is 5π6\frac{5\pi}{6}.

A

arg⁡z\arg z.

[3]
B

∣(1+i)z2∣|(1+i)z^2|.

[1]
Question 11
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The complex numbers ww and zz satisfy

wz=1−2i,z∗+2w=7−i\frac{w}{z}=1-2{\text{i}},\qquad z^*+2w=7-{\text{i}}

where z≠0z\ne0.

A

, if z=x+yiz=x+y{\text{i}}, then 3x+4y=73x+4y=7 and y−4x=−1y-4x=-1.

[3]
B

zz and ww in Cartesian form.

[3]
Question 12
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

On an Argand diagram, the point AA represents the complex number z=−3+4iz=-3+4{\text{i}}. The point BB represents the complex number ww, where

w=(2−i)zw=(2-{\text{i}})z
A

ww in Cartesian form.

[2]
B

fully the geometrical effect of multiplying by 2−i2-{\text{i}} on an Argand diagram.

[2]
C

the distance OBOB, where OO is the origin.

[1]

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Question 13
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let z=−4+6iz=-4+6{\text{i}}.

A

Write zz in Euler form reiθre^{{\text{i}}\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi.

[3]
B

z5z^5 in Cartesian form.

[3]
Question 14
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let

u=16(cos⁡2π3+isin⁡2π3)u=16\left(\cos\frac{2\pi}{3}+{\text{i}}\sin\frac{2\pi}{3}\right)
A

all values of u14u^{\frac{1}{4}} in modulus-argument form.

[4]
B

the argument of the root which lies in the third quadrant.

[1]
Question 15
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Two complex impedances are given by Z1=4+3iZ_1=4+3{\text{i}} and Z2=6−2iZ_2=6-2{\text{i}}. The combined impedance ZZ is defined by

1Z=1Z1+1Z2\frac{1}{Z}=\frac{1}{Z_1}+\frac{1}{Z_2}
A

ZZ in Cartesian form.

[3]
B

ZZ in the form reiθre^{{\text{i}}\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi.

[2]
Question 16
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let z=x+yiz=x+y{\text{i}}, where x,y∈Rx,y\in\mathbb{R}. The point representing zz on an Argand diagram satisfies both

∣z−4∣=∣z−2i∣|z-4|=|z-2{\text{i}}|

and

arg⁡(z−1−i)=π4\arg(z-1-{\text{i}})=\frac{\pi}{4}
A

the first condition gives y=2x−3y=2x-3.

[2]
B

zz.

[2]

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Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the equation z3=8iz^3=8i, where z∈Cz\in\mathbb{C}.

A

Write 8i8i in modulus-argument form.

[2]
B

the equation, giving your answers in Cartesian form.

[4]
Question 18
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let w=16cis⁡(2π3)w=16\operatorname{cis}\left(\frac{2\pi}{3}\right).

A

all values of w14w^{\frac{1}{4}} in modulus-argument form.

[5]
Question 19
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A monic quartic polynomial P(z)P(z) has real coefficients. Two of its roots are 1+2i1+2i and 3−i3-i.

A

the other two roots of P(z)P(z).

[2]
B

P(z)P(z) in expanded form.

[4]
Question 20
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta, where θ∈R\theta\in\mathbb{R}.

A

Expand z3z^3 in terms of cos⁡θ\cos\theta and sin⁡θ\sin\theta.

[2]
B

Use De Moivre's theorem to that cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta=4\cos^3\theta-3\cos\theta.

[3]

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Question 21
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The five roots of z5=1z^5=1 are zk=cis⁡(2πk5)z_k=\operatorname{cis}\left(\frac{2\pi k}{5}\right), where k=0,1,2,3,4k=0,1,2,3,4.

A

the sum of the five roots is 00.

[2]
B

cos⁡(2π5)+cos⁡(4π5)=−12\cos\left(\frac{2\pi}{5}\right)+\cos\left(\frac{4\pi}{5}\right)=-\frac{1}{2}.

[3]
Question 22
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

z3=−8+8iz^3=-8+8{\text{i}}
A

Write −8+8i-8+8{\text{i}} in modulus-argument form.

[2]
B

all solutions of the equation, giving your answers in modulus-argument form.

[4]
Question 23
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The polynomial

P(z)=z4−2z3+az2+bz+65P(z)=z^4-2z^3+az^2+bz+65

has real coefficients. Two of its roots are 2+3i2+3{\text{i}} and −1+2i-1+2{\text{i}}.

A

the other two roots of P(z)=0P(z)=0.

[1]
B

the values of aa and bb.

[4]
Question 24
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let z=x+yiz=x+y{\text{i}}, where x,y∈Rx,y\in\mathbb{R}. The complex number zz satisfies

∣z−(3+i)∣=2∣z+i∣|z-(3+{\text{i}})|=2|z+{\text{i}}|
A

the locus of zz has equation

3x2+3y2+6x+10y−6=03x^2+3y^2+6x+10y-6=0
[3]
B

the centre and radius of this locus.

[3]

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Question 25
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let z=x+yiz=x+yi, where x,y∈Rx,y\in\mathbb{R}, and define w=(1−i)z+2w=(1-i)z+2.

A
I.

Express ww in the form a+bia+bi, where a,b∈Ra,b\in\mathbb{R}, in terms of xx and yy.

[2]
II.

∣w∣2=(x+y+2)2+(y−x)2|w|^2=(x+y+2)^2+(y-x)^2.

[3]
B
I.

Given that ww is real, y=xy=x.

[1]
II.

Given also that ∣z−1∣=5|z-1|=\sqrt{5}, the possible values of zz.

[3]
C

The locus of zz is defined by ∣w∣=∣z∣|w|=|z|. the centre and radius of this locus in the Argand diagram.

[3]
Question 26
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let u=3+iu=\sqrt{3}+i, v=1−iv=1-i and q=uvq=\dfrac{u}{v}.

A
I.

Write uu and vv in modulus-argument form.

[2]
II.

write qq in modulus-argument form.

[2]
B
I.

q3q^3 in Cartesian form.

[3]
C

the smallest positive integer nn such that qnq^n is a positive imaginary number.

[3]
Question 27
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The transformation TT of the complex plane is defined by

T(z)=(1+3i)z+2−iT(z)=(1+\sqrt{3}{\text{i}})z+2-{\text{i}}.

The points AA, BB and CC represent the complex numbers 1+2i1+2{\text{i}}, 4+i4+{\text{i}} and 2−2i2-2{\text{i}} respectively.

A
I.

the modulus and principal argument of 1+3i1+\sqrt{3}{\text{i}}.

[2]
II.

fully the geometrical effect of multiplying a complex number by 1+3i1+\sqrt{3}{\text{i}}.

[2]
B

the complex numbers represented by T(A)T(A), T(B)T(B) and T(C)T(C), in Cartesian form.

[3]
C

the area of the triangle with vertices T(A)T(A), T(B)T(B) and T(C)T(C).

[3]
Question 28
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

z4=81(cos⁡2π3+isin⁡2π3)z^4=81\left(\cos\frac{2\pi}{3}+{\text{i}}\sin\frac{2\pi}{3}\right),

where z∈Cz\in\mathbb{C}.

A
I.

the modulus of each root.

[1]
II.

all four roots in modulus-argument form.

[5]
B

Give the roots in Cartesian form.

[4]
C

the area of the quadrilateral formed by joining the four roots in order around the origin.

[2]

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Question 29
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A monic quartic polynomial P(z)P(z) has real coefficients. One root of P(z)=0P(z)=0 is 32+2i\frac32+2{\text{i}}. Another root is −1-1, and the constant term of P(z)P(z) is −25-25.

A
I.

another non-real root of P(z)=0P(z)=0.

[1]
II.

the remaining real root.

[3]
B

P(z)P(z) in expanded form.

[4]
C

P(2)P(2).

[2]
Question 30
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Consider the equation
(z−1+i)4=16cis⁡(2π3),z∈C(z-1+i)^4=16\operatorname{cis}\left(\frac{2\pi}{3}\right),\qquad z\in\mathbb{C}

A
I.

Let ζ=z−1+i\zeta=z-1+i. all possible values of ζ\zeta in modulus-argument form.

[4]
II.

all possible values of zz in modulus-argument form translated into Cartesian form.

[1]
B
I.

one root of the original equation is real, and this root.

[2]
II.

If you did not obtain the real root in part (b)(i), use z=1+3z=1+\sqrt{3}. directly that this value satisfies the equation.

[1]
C

The four roots are represented by points in an Argand diagram. the area of the quadrilateral formed by joining these points in order of increasing argument of ζ\zeta.

[3]
Question 31
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A monic quartic polynomial P(z)P(z) has real coefficients. One root is 1+3i1+3i. The coefficient of z3z^3 is −6-6 and the constant term is −50-50.

A
I.

another root of P(z)P(z).

[1]
II.

Let the remaining two roots be real numbers rr and ss. r+s=4r+s=4 and rs=−5rs=-5.

[3]
B
I.

rr and ss.

[2]
II.

write P(z)P(z) as a product of real quadratic and linear factors.

[1]
C

P(z)P(z) in expanded form.

[3]
Question 32
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let 0<θ<π20<\theta<\dfrac{\pi}{2} and let
z=1+cos⁡2θ+isin⁡2θz=1+\cos2\theta+i\sin2\theta

A
I.

z=2cos⁡θcis⁡θz=2\cos\theta\operatorname{cis}\theta.

[3]
II.

the modulus and principal argument of zz.

[2]
B
I.

all cube roots of zz in modulus-argument form.

[3]
II.

If you did not obtain the roots in part (b)(i), use 2cos⁡θ3cis⁡(θ+2kπ3)\sqrt[3]{2\cos\theta}\operatorname{cis}\left(\dfrac{\theta+2k\pi}{3}\right), k=0,1,2k=0,1,2. the product of the three cube roots is zz.

[2]

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Question 33
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A transformation of the complex plane is defined by w=az+bw=az+b, where a,b∈Ca,b\in\mathbb{C} and a≠0a\ne0. The transformation maps 00 to 1+i1+i and maps 22 to 3+5i3+5i.

A
I.

aa and bb.

[3]
II.

geometrically the effect of multiplication by aa.

[1]
B
I.

The point zz moves on the circle ∣z−1∣=2|z-1|=2. the centre and radius of the image circle in the ww-plane.

[3]
II.

Write the Cartesian equation of the image circle, using w=X+Yiw=X+Yi.

[1]
C

the maximum possible value of Re⁡(w)\operatorname{Re}(w) for points on ∣z−1∣=2|z-1|=2.

[4]
Question 34
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let z=x+yiz=x+yi, where x,y∈Rx,y\in\mathbb{R}. The point representing zz in the Argand diagram satisfies
∣z−2∣=2∣z+1∣|z-2|=2|z+1|

A
I.

the locus has equation x2+4x+y2=0x^2+4x+y^2=0.

[3]
II.

the centre and radius of this locus.

[2]
B
I.

The point also satisfies arg⁡(z+1)=π4\arg(z+1)=\dfrac{\pi}{4}. y=x+1y=x+1 and the required restriction on xx.

[2]
II.

If you did not obtain the line in part (b)(i), use y=x+1y=x+1. zz exactly.

[3]
Question 35
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let a∈Ra\in\mathbb{R} and consider the quadratic equation
z2−2az+a2+16=0z^2-2az+a^2+16=0

A
I.

the roots are a+4ia+4i and a−4ia-4i.

[2]
II.

The root with positive imaginary part has modulus 55 and a>0a>0. aa.

[2]
B
I.

Using a=3a=3, the square roots of 3+4i3+4i.

[3]
C

the monic quartic polynomial with real coefficients whose roots are all the square roots of the two roots of the quadratic when a=3a=3.

[3]
Question 36
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A complex number z=x+yiz=x+y{\text{i}} is represented by a point PP on an Argand diagram. The point PP satisfies

∣z−(2+3i)∣=2|z-(2+3{\text{i}})|=2

and

arg⁡(z+1−i)=0.700\arg(z+1-{\text{i}})=0.700.

Angles are in radians.

A
I.

points on the ray may be written in the form z=−1+i+t(cos⁡0.700+isin⁡0.700)z=-1+{\text{i}}+t(\cos 0.700+{\text{i}}\sin 0.700), where t>0t>0.

[2]
II.

Substitute this form of zz into the circle equation to obtain a quadratic equation in tt.

[2]
B

the two possible values of zz.

[3]
C
I.

the distance between the two possible positions of PP.

[2]
II.

the smaller angle subtended at the origin by the two possible positions of PP.

[2]

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Question 37
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

In an alternating-current circuit, two impedances are connected in parallel. At angular frequency ω>0\omega>0 they are modelled by

Z1=6+0.5ωi,Z2=10−60ωiZ_1=6+0.5\omega{\text{i}},\qquad Z_2=10-\frac{60}{\omega}{\text{i}}.

The combined impedance ZZ satisfies

1Z=1Z1+1Z2\frac{1}{Z}=\frac{1}{Z_1}+\frac{1}{Z_2}.

A
I.

ZZ when ω=20\omega=20, in Cartesian form.

[3]
II.

Write this value of ZZ in Euler form reiθre^{{\text{i}}\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi.

[2]
B

ZZ is purely real when ω\omega satisfies

60ω100ω2+3600=0.5ω36+0.25ω2\frac{60\omega}{100\omega^2+3600}=\frac{0.5\omega}{36+0.25\omega^2}.

[3]
C

the value of ω\omega for which ZZ is purely real.

[3]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Consider the equation

(z−2+i)3=64(cos⁡5π6+isin⁡5π6)(z-2+{\text{i}})^3=64\left(\cos\frac{5\pi}{6}+{\text{i}}\sin\frac{5\pi}{6}\right).

A
I.

Let u=z−2+iu=z-2+{\text{i}}. the three possible values of uu in modulus-argument form.

[4]
II.

write the three solutions for zz in Cartesian form, correct to three significant figures.

[3]
B

the three solution points form an equilateral triangle and its area.

[3]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let z=x+yiz=x+y{\text{i}}, where x,y∈Rx,y\in\mathbb{R}. The complex number zz satisfies

z2+2z∗=11+8iz^2+2z^*=11+8{\text{i}}.

A
I.

xx and yy satisfy

x2−y2+2x=11,2xy−2y=8x^2-y^2+2x=11,\qquad 2xy-2y=8.

[3]
II.

xx satisfies x4−14x2+24x−27=0x^4-14x^2+24x-27=0.

[1]
B

the equation for zz, giving all solutions correct to three significant figures.

[5]
C

For the solution with positive imaginary part, the modulus and principal argument of zz.

[2]
Question 40
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The complex number u=3+iu=\sqrt{3}+i is used to generate a sequence of points P0,P1,P2,…P_0,P_1,P_2,\ldots on an Argand diagram. The point PnP_n represents the complex number zn=unz_n=u^n, where n∈Z≥0n\in\mathbb{Z}_{\ge 0}.

A
I.

Express uu in modulus-argument form.

[2]
II.

z3z_3 in Cartesian form.

[2]
B

the distance PnPn+1P_nP_{n+1} is 2n5−232^n\sqrt{5-2\sqrt{3}}.

[3]
C
I.

Let SN=z0+z1+z2+⋯+zNS_N=z_0+z_1+z_2+\cdots+z_N. SN=uN+1−1u−1S_N=\frac{u^{N+1}-1}{u-1}.

[2]
II.

the least value of NN such that ∣SN∣>1000|S_N|>1000. If you did not obtain the result in part (c)(i), use SN=uN+1−1u−1S_N=\frac{u^{N+1}-1}{u-1}.

[3]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Consider the equation

z5=32cis⁡ϕz^5=32\operatorname{cis}\phi

where 0≤ϕ<2π0\le \phi<2\pi. The five roots are represented by points on an Argand diagram.

A
I.

For ϕ=π2\phi=\frac{\pi}{2}, the five roots in modulus-argument form.

[3]
B

, for any value of ϕ\phi, the five roots form a regular pentagon with side length 4sin⁡(π5)4\sin\left(\frac{\pi}{5}\right).

[3]
C
I.

Let R0R_0 be one of the roots. the product of the distances from R0R_0 to the other four roots is 8080.

[3]
II.

the area of the pentagon and why it is independent of ϕ\phi. If you did not the regular pentagon result in part (b), use that the roots lie equally spaced on a circle of radius 22.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For t≥0t\ge 0, define

H(t)=1+it2−itH(t)=\frac{1+it}{2-it}

In an engineering interpretation, ∣H(t)∣|H(t)| represents a gain and arg⁡H(t)\arg H(t) represents a phase shift.

t

\

H(t)\

arg H(t) [rad]

0

0.5000

0.0000

0.5

0.5423

0.7086

1

0.6325

1.2490

1.5

0.7211

1.6263

2

0.7906

1.8925

3

0.8771

2.2318

4

0.9220

2.4330

A
I.

H(1)H(1) in Cartesian form.

[2]
II.

∣H(1)∣|H(1)|.

[1]
B
I.

∣H(t)∣2=1+t24+t2|H(t)|^2=\frac{1+t^2}{4+t^2}.

[2]
II.

the value of tt for which ∣H(t)∣=12|H(t)|=\frac{1}{\sqrt{2}}.

[2]
C
I.

arg⁡H(t)=arctan⁡t+arctan⁡(t2)\arg H(t)=\arctan t+\arctan\left(\frac{t}{2}\right) for t≥0t\ge 0.

[2]
II.

arg⁡H(t)=π3\arg H(t)=\frac{\pi}{3}. If you did not obtain part (c)(i), use tan⁡(arg⁡H(t))=3t21−t22\tan(\arg H(t))=\frac{\frac{3t}{2}}{1-\frac{t^2}{2}}.

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A transformation TT of the complex plane is defined by

T(z)=(1+i)z+2−iT(z)=(1+i)z+2-i

Repeated application of TT produces a sequence z0,z1,z2,…z_0,z_1,z_2,\ldots where zn+1=T(zn)z_{n+1}=T(z_n).

A
I.

T(3+i)T(3+i) in Cartesian form.

[2]
II.

the geometric effect of multiplying by 1+i1+i.

[2]
B
I.

the fixed point pp of TT, where T(p)=pT(p)=p.

[2]
II.

T(z)−p=(1+i)(z−p)T(z)-p=(1+i)(z-p).

[3]
C

Let z0=3+iz_0=3+i. the least value of nn such that ∣zn−p∣>100|z_n-p|>100. If you did not obtain the fixed point in part (b)(i), use p=1+2ip=1+2i.

[3]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a real parameter aa, consider the equation

(z−a)3=8(z-a)^3=8

where z∈Cz\in\mathbb{C}. The three roots are represented by points on an Argand diagram.

A
I.

For a=1a=1, the three roots in Cartesian form.

[3]
B

, for any real value of aa, the centroid of the three points representing the roots is the point representing aa.

[2]
C
I.

all real values of aa for which at least one root lies on the imaginary axis (has zero real part).

[3]
II.

that the area of the triangle formed by the three roots is independent of aa, and this area. If you did not obtain the roots in part (c)(i), use that the roots are obtained by translating the cube roots of 88 by aa.

[3]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

This question compares the values of z1/3z^{1/3} and z2/3z^{2/3} for complex numbers on the unit circle. Let

z=cis⁡t,qquad0≤t<2πz=\operatorname{cis}t,\\qquad 0\le t<2\pi
A
I.

all values of (−1)1/3(-1)^{1/3} in modulus-argument form.

[3]
B
I.

For general tt, all values of z1/3z^{1/3}.

[2]
II.

the three values of z2/3z^{2/3} form an equilateral triangle on the unit circle.

[2]
C

the area of the triangle formed by the three values of z2/3z^{2/3}, and why this area does not depend on tt.

[3]
Question 46
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta, where θ∈R\theta\in\mathbb{R}.

A
I.

Expand (cos⁡θ+isin⁡θ)5(\cos\theta+i\sin\theta)^5 using the binomial theorem.

[2]
II.

cos⁡5θ=16cos⁡5θ−20cos⁡3θ+5cos⁡θ\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta.

[3]
B
I.

Let α=π10\alpha=\dfrac{\pi}{10}. 16cos⁡4α−20cos⁡2α+5=016\cos^4\alpha-20\cos^2\alpha+5=0.

[3]
II.

the exact value of cos⁡π10\cos\dfrac{\pi}{10}.

[5]
Question 47
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let ω=cis⁡(2π5)\omega=\operatorname{cis}\left(\dfrac{2\pi}{5}\right).

A
I.

ω5=1\omega^5=1 and ω≠1\omega\ne1.

[2]
II.

1+ω+ω2+ω3+ω4=01+\omega+\omega^2+\omega^3+\omega^4=0.

[2]
B
I.

Divide the equation in part (a)(ii) by ω2\omega^2 and t=ω+ω−1t=\omega+\omega^{-1} satisfies t2+t−1=0t^2+t-1=0.

[3]
II.

2cos⁡2π52\cos\dfrac{2\pi}{5}.

[1]
C

the exact values of cos⁡2π5\cos\dfrac{2\pi}{5} and cos⁡4π5\cos\dfrac{4\pi}{5}.

[4]
Question 48
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For real α\alpha and β\beta, define cis⁡α=cos⁡α+isin⁡α\operatorname{cis}\alpha=\cos\alpha+i\sin\alpha.

A
I.

cis⁡αcis⁡β=cis⁡(α+β)\operatorname{cis}\alpha\operatorname{cis}\beta=\operatorname{cis}(\alpha+\beta).

[3]
II.

Use mathematical induction to De Moivre's theorem, (cis⁡θ)n=cis⁡(nθ)(\operatorname{cis}\theta)^n=\operatorname{cis}(n\theta), for all n∈Z+n\in\mathbb{Z}^+.

[3]
B
I.

Write 1+i1+i in modulus-argument form.

[1]
II.

(1+i)10(1+i)^{10} in Cartesian form.

[3]

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Question 49
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A sequence of complex numbers is defined by z0=0z_0=0 and
zn+1=(1+i)zn+1−iz_{n+1}=(1+i)z_n+1-i
for n≥0n\ge0.

A
I.

the fixed point α\alpha satisfying α=(1+i)α+1−i\alpha=(1+i)\alpha+1-i.

[2]
II.

zn+1−α=(1+i)(zn−α)z_{n+1}-\alpha=(1+i)(z_n-\alpha).

[2]
B
I.

zn=1+i−(1+i)n+1z_n=1+i-(1+i)^{n+1}.

[2]
II.

z3z_3.

[2]
C

If you did not obtain the expression in part (b)(i), use zn=1+i−(1+i)n+1z_n=1+i-(1+i)^{n+1}. the smallest positive integer nn such that zn−(1+i)z_n-(1+i) is a positive real number.

[3]
Question 50
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let z=cos⁡θ+isin⁡θz=\cos\theta+{\text{i}}\sin\theta, where θ∈R\theta\in\mathbb{R}.

A
I.

De Moivre's theorem for positive integers nn.

[1]
II.

De Moivre's theorem for positive integers by induction.

[4]
B

Use De Moivre's theorem to

cos⁡4θ=8cos⁡4θ−8cos⁡2θ+1\cos4\theta=8\cos^4\theta-8\cos^2\theta+1.

[4]
C

8cos⁡4θ−8cos⁡2θ+1=0.2008\cos^4\theta-8\cos^2\theta+1=0.200 for 0≤θ<2π0\le\theta<2\pi.

[4]
Question 51
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A sequence of complex numbers is defined by

z0=0,zn+1=azn+2−iz_0=0,\qquad z_{n+1}=az_n+2-{\text{i}},

where a=0.85e0.350ia=0.85e^{0.350{\text{i}}} and n∈Nn\in\mathbb{N}.

A
I.

the fixed point pp satisfying p=ap+2−ip=ap+2-{\text{i}}.

[3]
II.

zn=p(1−an)z_n=p(1-a^n) for n≥0n\ge0.

[2]
B

the least value of nn for which ∣zn−p∣<0.050|z_n-p|<0.050.

[4]
C

The total distance travelled by the first NN steps is DN=∑k=0N−1∣zk+1−zk∣D_N=\sum_{k=0}^{N-1}|z_{k+1}-z_k|. the least NN for which DND_N exceeds 90%90\% of its limiting value.

[3]
Question 52
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let AA and BB be the points representing 2+i2+{\text{i}} and −1+3i-1+3{\text{i}} respectively. A variable point PP represents the complex number zz.

A
I.

why arg⁡(z−(2+i)z−(−1+3i))=π2\arg\left(\frac{z-(2+{\text{i}})}{z-(-1+3{\text{i}})}\right)=\frac{\pi}{2} implies that ∠APB=π2\angle APB=\frac{\pi}{2}.

[2]
II.

the Cartesian equation of the locus satisfying this angle condition.

[3]
B

The point PP also satisfies ∣z∣=3|z|=3. the value of zz which satisfies both conditions and the stated argument condition.

[5]
C

the greatest possible value of ∣z∣|z| for a point on the circle found in part (a)(ii).

[3]

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Question 53
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let ω=cos⁡2π7+isin⁡2π7\omega=\cos\frac{2\pi}{7}+{\text{i}}\sin\frac{2\pi}{7}.

A
I.

all the roots of z7=1z^7=1.

[2]
II.

1+ω+ω2+⋯+ω6=01+\omega+\omega^2+\cdots+\omega^6=0.

[2]
B

Let S=ω+ω2+ω4S=\omega+\omega^2+\omega^4. S+S∗=−1S+S^*=-1 and SS∗=2SS^*=2.

[5]
C

SS in Cartesian form.

[2]
D

the exact value of sin⁡2π7+sin⁡4π7+sin⁡8π7\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}.

[2]
Question 54
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two fixed points AA and BB on an Argand diagram represent the complex numbers 2+i2+i and −1+3i-1+3i respectively. For a variable complex number z≠−1+3iz\ne -1+3i, define

w=z−(2+i)z−(−1+3i)w=\frac{z-(2+i)}{z-(-1+3i)}
A
I.

ww when z=4+5iz=4+5i, giving your answer in Cartesian form.

[3]
II.

∣w∣|w| for this value of zz.

[1]
B
I.

the condition ∣w∣=1|w|=1 represents a straight line, and its equation in the form ax+by+c=0ax+by+c=0, where z=x+yiz=x+yi.

[3]
II.

the condition ∣w∣=2|w|=2 represents a circle, and its centre and radius.

[4]
C

zz when ∣w∣=2|w|=2 and arg⁡w=π3\arg w=\frac{\pi}{3}. If you did not obtain a circle in part (b)(ii), continue using the definition of ww.

[3]
Question 55
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For r>1r>1 and 0<θ<π20<\theta<\frac{\pi}{2}, consider the four complex numbers

rcis⁡θ,rcis⁡(−θ),r−1cis⁡θ,r−1cis⁡(−θ)r\operatorname{cis}\theta,\quad r\operatorname{cis}(-\theta),\quad r^{-1}\operatorname{cis}\theta,\quad r^{-1}\operatorname{cis}(-\theta)

They are represented by four points on an Argand diagram.

A
I.

For r=2r=2 and θ=π3\theta=\frac{\pi}{3}, the four complex numbers in Cartesian form.

[2]
B

Let P(z)P(z) be the monic quartic polynomial having the four complex numbers in the stem as roots.

P(z)=z4−2(r+r−1)cos⁡θ z3+(r2+r−2+4cos⁡2θ)z2−2(r+r−1)cos⁡θ z+1P(z)=z^4-2\left(r+r^{-1}\right)\cos\theta\,z^3+\left(r^2+r^{-2}+4\cos^2\theta\right)z^2-2\left(r+r^{-1}\right)\cos\theta\,z+1
[4]
C
I.

For r=2r=2 and θ=π3\theta=\frac{\pi}{3}, the area of the quadrilateral formed by joining the four points in order around the boundary.

[3]
II.

a formula for the area of the quadrilateral in terms of rr and θ\theta. If you did not obtain the area in part (c)(i), use the same symmetric quadrilateral described in the stem.

[3]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let z=cis⁡θz=\operatorname{cis}\theta, where θ∈R\theta\in\mathbb{R}. Define Cn(x)C_n(x) by

zn+z−n=2Cn(x),x=cos⁡θz^n+z^{-n}=2C_n(x),\qquad x=\cos\theta

This question investigates a recurrence for the functions CnC_n.

A
I.

C2(x)=2x2−1C_2(x)=2x^2-1.

[2]
II.

, for n≥1n\ge 1, Cn+1(x)=2xCn(x)−Cn−1(x)C_{n+1}(x)=2xC_n(x)-C_{n-1}(x).

[3]
B
I.

Use the recurrence to C4(x)C_4(x).

[3]
C

, or otherwise, cos⁡4θ=cos⁡θ\cos 4\theta=\cos\theta for 0≤θ<2π0\le \theta<2\pi. If you did not obtain C4(x)C_4(x), use cos⁡4θ=8cos⁡4θ−8cos⁡2θ+1\cos 4\theta=8\cos^4\theta-8\cos^2\theta+1.

[4]

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Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A monic quartic polynomial has real coefficients and is reciprocal, meaning that if z≠0z\ne0 is a root then 1z\frac{1}{z} is also a root. One root is

α=2cis⁡(π4)\alpha=2\operatorname{cis}\left(\frac{\pi}{4}\right)
A
I.

the other three roots.

[2]
II.

the polynomial in the form z4+pz3+qz2+pz+1z^4+pz^3+qz^2+pz+1.

[4]
B

if P(z)=z4+pz3+qz2+pz+1P(z)=z^4+pz^3+qz^2+pz+1, then P(z)=0P(z)=0 implies P(1z)=0P\left(\frac{1}{z}\right)=0 for z≠0z\ne0.

[3]
C

For roots of the form 2cis⁡θ2\operatorname{cis}\theta, 2cis⁡(−θ)2\operatorname{cis}(-\theta), 12cis⁡θ\frac{1}{2}\operatorname{cis}\theta, 12cis⁡(−θ)\frac{1}{2}\operatorname{cis}(-\theta), the least possible value of the coefficient qq.

[2]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a positive integer nn and real number θ\theta, define the phasor sum

Sn(θ)=1+cis⁡θ+cis⁡2θ+⋯+cis⁡(n−1)θS_n(\theta)=1+\operatorname{cis}\theta+\operatorname{cis}2\theta+\cdots+\operatorname{cis}(n-1)\theta
A
I.

, when cis⁡θ≠1\operatorname{cis}\theta\ne1,

Sn(θ)=1−cis⁡(nθ)1−cis⁡θS_n(\theta)=\frac{1-\operatorname{cis}(n\theta)}{1-\operatorname{cis}\theta}
[3]
II.

, when cis⁡θ≠1\operatorname{cis}\theta\ne1,

∣Sn(θ)∣=∣sin⁡(nθ2)sin⁡(θ2)∣|S_n(\theta)|=\left|\frac{\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\right|
[3]
B

Evaluate S6(π5)S_6\left(\frac{\pi}{5}\right) in Cartesian form. If you did not obtain the formula in part (a), use the geometric series formula.

[2]
C

that the sum of all nn roots of zn=1z^n=1 is zero for n>1n>1.

[2]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let zz be a non-zero complex number and define

w=z+1zw=z+\frac{1}{z}

This question investigates the special case in which zz lies on the unit circle.

A
I.

If z=cis⁡θz=\operatorname{cis}\theta, z+1z=2cos⁡θz+\frac{1}{z}=2\cos\theta.

[2]
II.

zn+z−n=2cos⁡(nθ)z^n+z^{-n}=2\cos(n\theta) for any positive integer nn.

[2]
B

Suppose ww is real and −2<w<2-2<w<2. that the two roots of z2−wz+1=0z^2-wz+1=0 are complex conjugates lying on the unit circle.

[4]
C
I.

For w=1w=1, the two roots of z2−wz+1=0z^2-wz+1=0 in modulus-argument form.

[2]
II.

z6+z−6z^6+z^{-6} for either root. If you did not obtain the roots in part (c)(i), use z=cis⁡(π3)z=\operatorname{cis}\left(\frac{\pi}{3}\right).

[1]
Question 60
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For z≠3−iz\ne3-{\text{i}}, define

w=z−(1+i)z−(3−i)w=\frac{z-(1+{\text{i}})}{z-(3-{\text{i}})}.

Let z=x+yiz=x+y{\text{i}}.

A
I.

ww when z=4+2iz=4+2{\text{i}}.

[2]
II.

an expression for zz in terms of ww.

[2]
B

the condition ∣w∣=1|w|=1 corresponds to the line x−y−2=0x-y-2=0.

[3]
C

the condition arg⁡w=π4\arg w=\frac{\pi}{4} leads to

x2+y2−2x+2y−2=0x^2+y^2-2x+2y-2=0,

with a suitable restriction on the arc.

[4]
D

the value of zz for which both ∣w∣=1|w|=1 and arg⁡w=π4\arg w=\frac{\pi}{4}.

[3]

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Binomial Theorem

Counting Principles