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Complex Numbers

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

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Question 1
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

Let z=3−4i1+2iz=\dfrac{3-4i}{1+2i} and w=2+iw=2+i.

A

Express zz in Cartesian form.

[3]
B

zwzw.

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

Consider the expression 3i47−2i263i^{47}-2i^{26} and the quadratic equation x2−4x+13=0x^2-4x+13=0.

A

Express 3i47−2i263i^{47}-2i^{26} in Cartesian form.

[2]
B

the quadratic equation, giving the solutions in Cartesian form.

[3]
Question 3
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

Let z=−5+53iz=-5+5\sqrt3i and w=2e−iπ/6w=2e^{-i\pi/6}.

A

Express zz in exponential form, giving an exact argument in the interval −π<θ≤π-\pi<\theta\leq\pi.

[3]
B

Express ww in exact Cartesian form.

[2]
Question 4
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

A complex number zz has modulus 66 and principal argument −2π3-\dfrac{2\pi}{3}.

A

Write zz in exact Cartesian form.

[2]
B

Write the conjugate z‾\overline z in exact exponential form.

[1]
C

the exact value of z+z‾z+\overline z.

[1]

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Question 5
HL • Paper 1
Medium
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HL • Paper 1
Medium
Calculator Permitted

The complex number zz is given by z=k+(k−2)iz=k+(k-2)i, where k∈Rk\in\mathbb{R}. It is known that ∣z∣=25|z|=2\sqrt5 and Re⁡(z)>0\operatorname{Re}(z)>0.

A

the value of kk.

[3]
B

the principal argument of zz.

[2]
Question 6
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Let z1=−2+5iz_1=-2+5i, z2=4−iz_2=4-i and w=z1−z2w=z_1-z_2.

A

ww in Cartesian form.

[1]
B

On an Argand diagram, and the points representing z1z_1, z2z_2 and ww.

[2]
C

∣w∣|w| and the principal argument of ww, giving exact values.

[3]
Question 7
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The function ff is defined by f(x)=2x2−8x+13f(x)=2x^2-8x+13, where x∈Rx\in\mathbb{R}.

A

the discriminant of the equation f(x)=0f(x)=0.

[2]
B

the polynomial equation 2x2−8x+13=02x^2-8x+13=0 over C\mathbb{C}, giving the solutions in exact Cartesian form.

[2]
C

how the solutions in part (b) are consistent with the graph of y=f(x)y=f(x) having no xx-intercepts.

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

Let z=4cis⁡(5π6)z=4\operatorname{cis}\left(\dfrac{5\pi}{6}\right) and w=2cis⁡(−π3)w=2\operatorname{cis}\left(-\dfrac{\pi}{3}\right).

A

zwzw, giving the answer in exact Cartesian form.

[2]
B

zw\dfrac{z}{w}, giving the answer in exact Cartesian form with principal argument.

[3]

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

The complex number zz is given by z=3+iz=\sqrt3+i.

A

Write zz in exact modulus–argument form.

[2]
B

z5z^5 in exact Cartesian form.

[2]
C

the exact value of z12z^{12}.

[1]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Let z=2−3iz=2-3i.

A

Using technology, z6z^6 in Cartesian form.

[2]
B

that ∣z6∣=∣z∣6|z^6|=|z|^6.

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

A transformation of the complex plane is defined by w=αzw=\alpha z, where α=1+3i\alpha=1+\sqrt3i. The point PP represents the complex number z=−1+2iz=-1+2i, and its image is the point QQ representing ww.

A

Express α\alpha in exact modulus–argument form.

[2]
B

fully the geometric transformation that maps PP to QQ.

[2]
C

ww in exact Cartesian form.

[2]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Two alternating voltages, measured in volts, are given by
V1(t)=8cos⁡(12t+0.4),V2(t)=11cos⁡(12t−0.9)V_1(t)=8\cos(12t+0.4),\qquad V_2(t)=11\cos(12t-0.9)
where tt is measured in seconds. The total voltage is V(t)=V1(t)+V2(t)V(t)=V_1(t)+V_2(t).

A

an expression for V(t)V(t) in the form Rcos⁡(12t+ϕ)R\cos(12t+\phi), where R>0R>0 and −π<ϕ≤π-\pi<\phi\leq\pi.

[4]
B

the maximum total voltage.

[1]
C

the first time t≥0t\geq0 at which the maximum total voltage occurs.

[1]

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Question 13
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Three water waves have heights
h1(t)=4sin⁡(5t),h2(t)=3sin⁡(5t+π2),h3(t)=2sin⁡(5t+π)h_1(t)=4\sin(5t),\qquad h_2(t)=3\sin\left(5t+\frac{\pi}{2}\right),\qquad h_3(t)=2\sin(5t+\pi)
where height is measured in centimetres and tt in seconds. Their combined height is h(t)=h1(t)+h2(t)+h3(t)h(t)=h_1(t)+h_2(t)+h_3(t).

A

Express h(t)h(t) in the form Rsin⁡(5t+ϕ)R\sin(5t+\phi), where R>0R>0 and 0<ϕ<π20<\phi<\dfrac{\pi}{2}.

[4]
B

the maximum combined height.

[1]
C

the first time t>0t>0 at which the combined height is zero.

[1]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A complex number zz satisfies
z+z‾=8andzz‾=25z+\overline z=8\qquad\text{and}\qquad z\overline z=25

A

all possible values of zz in Cartesian form.

[3]
B

the principal argument of each possible value of zz.

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

An input signal has complex amplitude P=6−8iP=6-8i. A filter multiplies this amplitude by the transfer factor H=1+i2H=\dfrac{1+i}{2}. The output complex amplitude is Q=HPQ=HP.

A

Express PP in exponential form, giving the modulus and argument to three significant figures.

[2]
B

QQ in exact Cartesian form.

[2]
C

the amplitude and phase shift represented by QQ.

[1]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The points AA and BB on an Argand diagram represent zA=3+iz_A=3+i and zB=−1+4iz_B=-1+4i, respectively. Multiplication by a complex number qq maps AA to BB, so that zB=qzAz_B=qz_A.

A

qq in Cartesian form.

[3]
B

the modulus and principal argument of qq.

[2]
C

multiplication by qq as a geometric transformation of the Argand diagram.

[1]

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Question 17
HL • Paper 2
Medium
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HL • Paper 2
Medium
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Let z=−1+3iz=-1+\sqrt3i and define wn=znw_n=z^n for positive integers nn.

A
I.

Express zz in exact modulus–argument form.

[2]
II.

w5w_5 in exact Cartesian form.

[3]
B
I.

the least positive integer nn for which wnw_n is a positive real number.

[2]
II.

the least positive integer mm for which wn+mw_{n+m} has the same argument as wnw_n for every positive integer nn, and your answer.

[2]
Question 18
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Three actuators produce vibrations with the same angular frequency. Their phasors are P1=9e0.35iP_1=9e^{0.35i}, P2=6e−0.80iP_2=6e^{-0.80i} and P3=AeiθP_3=Ae^{i\theta}, where A>0A>0 and −π<θ≤π-\pi<\theta\leq\pi. The resultant phasor is P=P1+P2+P3P=P_1+P_2+P_3.

A
I.

Express P1+P2P_1+P_2 in Cartesian form.

[2]
II.

Write P1+P2P_1+P_2 in the form ReiϕRe^{i\phi}, where R>0R>0 and −π<ϕ≤π-\pi<\phi\leq\pi.

[2]
III.

write the vibration produced by the first two actuators in the form Rcos⁡(ωt+ϕ)R\cos(\omega t+\phi).

[2]
B

AA and θ\theta so that the three vibrations cancel completely. If you did not obtain an answer to part (a)(ii), use P1+P2=12.6932e−0.0961iP_1+P_2=12.6932e^{-0.0961i}.

[4]
Question 19
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A sequence of points on an Argand diagram is represented by z0=5−2iz_0=5-2i and
zn+1=qe0izn,q=0.82eiπ/7z_{n+1}=qe^{0i}z_n,\qquad q=0.82e^{i\pi/7}
for n≥0n\geq0.

A
I.

the scale factor and angle of rotation that map znz_n to zn+1z_{n+1}.

[2]
II.

Express z0z_0 in exponential form, giving numerical values to three significant figures.

[3]
B
I.

zn=5.39(0.82)nei(−0.381+nπ/7)z_n=5.39(0.82)^ne^{i(-0.381+n\pi/7)}, with numerical values given to three significant figures.

[2]
II.

z14z_{14} in Cartesian form.

[2]
III.

why z14z_{14} lies on the same ray from the origin as z0z_0 but is closer to the origin.

[1]
Question 20
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The equation
z2−6z+k=0z^2-6z+k=0
where k∈Rk\in\mathbb R, has two non-real roots. One root is denoted by α\alpha, and ∣α∣=5|\alpha|=5.

A
I.

why the other root is α‾\overline\alpha.

[1]
II.

Using the sum of the roots, Re⁡(α)\operatorname{Re}(\alpha).

[2]
III.

both possible values of α\alpha.

[2]
B
I.

kk.

[2]
II.

how the graph of y=x2−6x+ky=x^2-6x+k is consistent with the roots being non-real.

[2]

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Question 21
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A transformation TT of the complex plane is defined by T(z)=qzT(z)=qz. The point represented by 2−3i2-3i is mapped to the point represented by −4+7i-4+7i.

A
I.

qq in Cartesian form.

[3]
II.

the modulus and principal argument of qq.

[3]
B

fully the geometric effect of TT and the image of the point represented by 1+4i1+4i.

[4]
Question 22
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A complex-valued calculation is defined by
Q=(3−2i)8(1+i)5Q=\frac{(3-2i)^8}{(1+i)^5}

A
I.

Using technology, (3−2i)8(3-2i)^8 in Cartesian form.

[2]
II.

Express (1+i)5(1+i)^5 in exact Cartesian form.

[3]
B
I.

QQ in Cartesian form.

[3]
II.

numerically that ∣Q∣=∣3−2i∣8∣1+i∣5|Q|=\dfrac{|3-2i|^8}{|1+i|^5}.

[2]
Question 23
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The horizontal displacement, in millimetres, of a platform is modelled by
x(t)=5cos⁡(8t)+7cos⁡(8t+1.10)+4cos⁡(8t−0.65)x(t)=5\cos(8t)+7\cos\left(8t+1.10\right)+4\cos\left(8t-0.65\right)
where tt is measured in seconds.

A
AI.

the phasor corresponding to each of the three terms.

[2]
AII.

their sum in Cartesian form.

[2]
AIII.

Express the sum in exponential form.

[2]
B
BI.

express x(t)x(t) as a single cosine function.

[2]
BII.

the first time t>0t>0 at which the platform reaches its maximum positive displacement.

[3]
Question 24
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A quadratic function is given by f(x)=2x2+px+qf(x)=2x^2+px+q, where p,q∈Rp,q\in\mathbb R. The equation f(x)=0f(x)=0 has one root −1+3i-1+3i.

A
AI.

the other root and your answer.

[2]
AII.

pp and qq.

[4]
B
BI.

the coordinates of the vertex of the graph y=f(x)y=f(x).

[2]
BII.

the relationship between the vertex and the real and imaginary parts of the two roots.

[2]

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Question 25
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Let
z=3cis⁡(5π12),w=4cis⁡(−π8)z=3\operatorname{cis}\left(\frac{5\pi}{12}\right),\qquad w=4\operatorname{cis}\left(-\frac{\pi}{8}\right)

A
AI.

Express zwzw in exact modulus–argument form.

[2]
AII.

Express z3w2\dfrac{z^3}{w^2} in exact modulus–argument form, with principal argument.

[4]
B
BI.

express z3w2\dfrac{z^3}{w^2} in exact Cartesian form.

[2]
BII.

geometrically the effect of multiplying any complex number by z3w2\dfrac{z^3}{w^2}.

[2]
Question 26
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A point in an animation is represented by z0=5e−0.2iz_0=5e^{-0.2i}. Each frame multiplies its position by α=1.2e0.35i\alpha=1.2e^{0.35i}, so that zn=αnz0z_n=\alpha^nz_0 after nn frames.

A
I.

Express α\alpha in Cartesian form, giving its components to three significant figures.

[2]
II.

geometrically the effect of one frame on a point in the complex plane.

[2]
B

the first value of nn for which ∣zn∣>12|z_n|>12, and the corresponding position in Cartesian form.

[4]
Question 27
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Three synchronized seismic signals are
s1(t)=6cos⁡(4t+0.4),s2(t)=8cos⁡(4t+1.7),s3(t)=5cos⁡(4t−0.8)s_1(t)=6\cos(4t+0.4),\quad s_2(t)=8\cos(4t+1.7),\quad s_3(t)=5\cos(4t-0.8)
where displacement is measured in millimetres and tt in seconds.

A
I.

a complex phasor representing each signal.

[2]
II.

the resultant phasor in Cartesian form.

[2]
B

Express the resultant as s(t)=Rcos⁡(4t+ϕ)s(t)=R\cos(4t+\phi) and the duration in each cycle for which s(t)>9 mms(t)>9\ \text{mm}.

[4]
Question 28
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two loudspeakers produce signals
L1(t)=10cos⁡(6t+ϕ),L2(t)=10cos⁡(6t−ϕ)L_1(t)=10\cos(6t+\phi),\qquad L_2(t)=10\cos(6t-\phi)
where 0<ϕ<π20<\phi<\frac{\pi}{2}. The measured amplitude of their combined signal is 1313. Here, tt is measured in seconds, so the angular frequency is 6 rad s−16\ \text{rad s}^{-1}.

A
I.

using phasors that the resultant phasor is real and has modulus 20cos⁡ϕ20\cos\phi.

[2]
II.

ϕ\phi.

[2]
B

A third signal L3(t)=6cos⁡(6t+π2)L_3(t)=6\cos(6t+\frac{\pi}{2}) is added. Express the new total signal as Rcos⁡(6t+θ)R\cos(6t+\theta) and the first time t>0t>0 at which it reaches its maximum.

[4]

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Question 29
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The vertices of a regular hexagonal plate are represented by
zk=3ekπi/3,k=0,1,…,5z_k=3e^{k\pi i/3},\qquad k=0,1,\ldots,5
A manufacturing process maps each vertex to wk=αzkw_k=\alpha z_k, where α=1.5eiπ/4\alpha=1.5e^{i\pi/4}.

A
I.

fully the transformation from zkz_k to wkw_k.

[2]
II.

w2w_2 in exact Cartesian form.

[2]
B

the exact area of the original plate and the area of the transformed plate.

[4]
Question 30
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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A resultant signal is measured as
V(t)=10cos⁡(5t+0.4)V(t)=10\cos(5t+0.4)
One known component is V1(t)=8cos⁡(5t)V_1(t)=8\cos(5t). A second component is V2(t)=Acos⁡(5t+ϕ)V_2(t)=A\cos(5t+\phi), where A>0A>0 and −π<ϕ≤π-\pi<\phi\leq\pi. The variable tt is measured in seconds.

A
I.

Write an equation relating the three phasors.

[2]
II.

Express the phasor of V2V_2 in Cartesian form.

[2]
B

AA and ϕ\phi. the first time t>0t>0 at which the resultant signal is zero.

[4]
Question 31
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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The points AA, BB and CC on an Argand diagram represent the complex numbers a=1+2ia=1+2i, b=6+ib=6+i and c=4+5ic=4+5i, respectively. A fourth point DD is chosen so that ABCDABCD is a parallelogram, with the vertices in this order.

A
I.

Using complex-number addition, the complex number dd represented by DD.

[3]
II.

the lengths ABAB and BCBC.

[3]
B
I.

the angle ABCABC is 0.9100.910 radians, correct to three significant figures.

[3]
II.

classify the parallelogram as precisely as possible, justifying your answer.

[2]
Question 32
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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A logo is generated from the complex number z0=3e0.20iz_0=3e^{0.20i}. Successive vertices are defined by
zn+1=(1.05e0.60i)znz_{n+1}=\left(1.05e^{0.60i}\right)z_n
The centre of the logo is the origin.

A
I.

the transformation from znz_n to zn+1z_{n+1}.

[2]
II.

an expression for znz_n in exponential form.

[3]
B
I.

z10z_{10} in Cartesian form.

[2]
II.

The designer claims that z10z_{10} is almost on the same ray from the origin as z0z_0. the smaller angle between the two rays and on the claim.

[3]

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Question 33
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Let z=a+biz=a+bi, where a,b∈Ra,b\in\mathbb R, b≠0b\ne0 and z≠−1z\ne-1. Define
w=z−1z+1w=\frac{z-1}{z+1}

A
I.

Express ww in Cartesian form in terms of aa and bb.

[4]
II.

if ∣z∣=1|z|=1, then ww is purely imaginary.

[2]
B

For z=cos⁡(1.20)+isin⁡(1.20)z=\cos(1.20)+i\sin(1.20), ww in Cartesian form and the result from part (a)(ii).

[4]
Question 34
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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Four speakers emit tones of the same angular frequency. Their complex amplitudes are shown in the table. The total complex amplitude is the sum of the four amplitudes.

Speaker

Complex amplitude

Speaker 1

10e0i10e^{0i}

Speaker 2

8e0.90i8e^{0.90i}

Speaker 3

6e−1.40i6e^{-1.40i}

Speaker 4

5e2.20i5e^{2.20i}

A

Part (a)

I.

the total complex amplitude in Cartesian form.

[3]
II.

Express PP in the form ReiϕRe^{i\phi}.

[3]
B

Part (b)

I.

Write the resultant tone as s(t)=Rcos⁡(ωt+ϕ)s(t)=R\cos(\omega t+\phi).

[2]
II.

A fifth speaker is added so that the total amplitude becomes 2020 with phase shift zero. the fifth speaker's complex amplitude in Cartesian form.

[2]
III.

the amplitude and phase shift of the fifth speaker.

[2]
Question 35
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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Two alternating currents are modelled by
I1(t)=14cos⁡(50t+α),I2(t)=10cos⁡(50t−0.40)I_1(t)=14\cos(50t+\alpha),\qquad I_2(t)=10\cos(50t-0.40)
where current is measured in amperes. It is known that the resultant current has phase shift zero and a positive amplitude.

A
I.

Write an equation involving imaginary parts that must be satisfied by α\alpha.

[2]
II.

Given that 0<α<π20<\alpha<\frac{\pi}{2}, α\alpha.

[2]
III.

the resultant amplitude.

[2]
B
I.

write the resultant current as a single cosine function.

[1]
II.

A third current is to reduce the resultant amplitude to 55 amperes while retaining phase shift zero. one possible phasor for this third current and its phase.

[4]
Question 36
HL • Paper 2
Hard
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HL • Paper 2
Hard
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On an Argand diagram, the points OO, AA and BB represent 00, a=4+ia=4+i and b=−2+5ib=-2+5i, respectively. A transformation T(z)=qzT(z)=qz maps AA to BB.

A
I.

qq in Cartesian form.

[3]
II.

the exact scale factor of TT and its angle of rotation to three significant figures.

[4]
B
I.

The midpoint of ABAB represents the complex number mm. mm.

[2]
II.

T(m)T(m) in Cartesian form.

[2]
III.

why T(m)T(m) is the midpoint of T(a)T(a) and T(b)T(b).

[1]

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Question 37
HL • Paper 3
Hard
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HL • Paper 3
Hard
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Two vibration sources have phasors P1=12e0.2iP_1=12e^{0.2i} and P2=9e−1.1iP_2=9e^{-1.1i}. A controllable source has phasor C=5eθiC=5e^{\theta i}, where −π<θ≤π-\pi<\theta\leq\pi.

A
I.

P=P1+P2P=P_1+P_2 in Cartesian form.

[2]
II.

the modulus and principal argument of PP.

[2]
B

the value of θ\theta that minimizes ∣P+C∣|P+C|, and this minimum modulus.

[4]
Question 38
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For 0≤t<2π0\leq t<2\pi, consider the quadratic equation
x2−2(1+2cos⁡t)x+(5+4cos⁡t)=0x^2-2(1+2\cos t)x+(5+4\cos t)=0

A
I.

the discriminant is −16sin⁡2t-16\sin^2t.

[2]
II.

For 0<t<π0<t<\pi, write the root with positive imaginary part in Cartesian form.

[2]
B

For part (b), take z=1+2cos⁡t+2isin⁡tz=1+2\cos t+2i\sin t as the continuous continuation of the root identified in part (ii). the locus of zz and the two values of tt for which Re⁡(z)=Im⁡(z)\operatorname{Re}(z)=\operatorname{Im}(z).

[4]
Question 39
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A point moves by repeated rotations about the fixed point c=2−ic=2-i. Its positions satisfy
zn+1=c+q(zn−c),q=e2πi/7z_{n+1}=c+q(z_n-c),\qquad q=e^{2\pi i/7}
where z0=5+iz_0=5+i.

A
I.

z1−cz_1-c in terms of qq.

[2]
II.

zn=c+(3+2i)qnz_n=c+(3+2i)q^n.

[2]
B

z3z_3 in Cartesian form and why z7=z0z_7=z_0.

[4]
Question 40
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two beacons are represented on an Argand diagram by a=−2+3ia=-2+3i and b=6+ib=6+i. A receiver at z=x+yiz=x+yi is 55 units from beacon AA and 17\sqrt{17} units from beacon BB.

A
I.

two equations satisfied by xx and yy.

[2]
II.

y=4x−8y=4x-8.

[2]
B

Given that the receiver has positive imaginary part, zz, its principal argument and the area of triangle ABZABZ.

[4]

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Question 41
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A coded signal is multiplied repeatedly by q=1.04e0.28iq=1.04e^{0.28i}. Starting with z0=1z_0=1, the signal after nn stages is zn=qnz_n=q^n. Only stages 1≤n≤351\leq n\leq35 are available.

A
I.

Write znz_n in exponential form.

[2]
II.

∣z20∣|z_{20}| and a principal argument of z20z_{20}.

[2]
B

the value of nn for which znz_n is closest in direction to the negative real axis. the corresponding modulus.

[4]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let z=−3+2iz=-3+2i and define Sn=zn+z‾ nS_n=z^n+\overline z^{\,n} for non-negative integers nn.

A
I.

zz and z‾\overline z are roots of x2+6x+13=0x^2+6x+13=0.

[2]
II.

S0S_0, S1S_1 and S2S_2.

[2]
B

Sn=−6Sn−1−13Sn−2S_n=-6S_{n-1}-13S_{n-2} for n≥2n\geq2, and S8S_8.

[4]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A camera position is updated by
zn+1=qzn+(1−q)Lz_{n+1}=qz_n+(1-q)L
where q=0.8eiπ/6q=0.8e^{i\pi/6}, L=2+3iL=2+3i and z0=8z_0=8.

A
I.

zn+1−L=q(zn−L)z_{n+1}-L=q(z_n-L).

[2]
II.

an expression for znz_n.

[2]
B

the first value of nn for which the camera is less than 0.50.5 units from LL, and znz_n at this stage.

[4]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A monic quadratic with real coefficients has vertex (4,9)(4,9) and passes through (0,25)(0,25). Let zz be its root with positive imaginary part and define
w=z−1z+1w=\frac{z-1}{z+1}

Component

Key data

Real graph

Vertex (4,9); y-intercept (0,25); sample points (2,13) and (6,13); opens upward.

Argand diagram

Conjugate roots symmetric about the real axis; the upper root is z.

A
I.

the equation of the quadratic.

[2]
II.

zz and why the real graph has no xx-intercepts.

[2]
B

ww in Cartesian and exponential forms. the first positive integer nn for which ∣wn∣<0.1|w^n|<0.1.

[4]

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Question 45
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Let
u=e2πi/5u=e^{2\pi i/5}
The complex numbers 1,u,u2,u3,u41,u,u^2,u^3,u^4 represent five equally spaced points on the unit circle.

A

Part (a)

I.

u5=1u^5=1 and u≠1u\ne1.

[2]
II.

Using the identity u5−1=(u−1)(u4+u3+u2+u+1)u^5-1=(u-1)(u^4+u^3+u^2+u+1), that
1+u+u2+u3+u4=01+u+u^2+u^3+u^4=0

[2]
III.

the result geometrically on the Argand diagram.

[2]
B

Part (b)

I.

the exact value of u+u4u+u^4 in terms of u2+u3u^2+u^3.

[1]
II.

u+u4=2cos⁡(2π5)u+u^4=2\cos\left(\frac{2\pi}{5}\right) and u2+u3=2cos⁡(4π5)u^2+u^3=2\cos\left(\frac{4\pi}{5}\right).

[3]
III.

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

[2]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A transformation is defined by
T(z)=eiπ/3z‾+1−3iT(z)=e^{i\pi/3}\overline z+1-\sqrt3i

A
I.

T(3+2i)T(3+2i) in exact Cartesian form.

[2]
II.

T(T(z))=zT(T(z))=z.

[2]
B

the locus of fixed points of TT and the transformation geometrically.

[4]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A non-real complex number zz satisfies
z+1z=sz+\frac1z=s
where ss is real and −2<s<2-2<s<2.

A
I.

Writing z=reiθz=re^{i\theta}, r=1r=1.

[2]
II.

zz is a root of x2−sx+1=0x^2-sx+1=0.

[2]
B

For s=1.2s=1.2, the two possible values of zz. For the value with positive imaginary part, the first positive integer nn for which Re⁡(zn)<0\operatorname{Re}(z^n)<0.

[4]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two equal-amplitude control signals are represented by the phasors
P1=Aeiα,P2=AeiβP_1=Ae^{i\alpha},\qquad P_2=Ae^{i\beta}
where A>0A>0 and ∣α−β∣<π|\alpha-\beta|<\pi.

A
I.

Factorize P1+P2P_1+P_2 to its argument is the mean of α\alpha and β\beta.

[2]
II.

the modulus of the resultant.

[2]
B

For A=15A=15, the resultant has modulus 1818 and argument 0.70.7. all possible ordered pairs (α,β)(\alpha,\beta) satisfying −π<α,β≤π-\pi<\alpha,\beta\leq\pi.

[4]

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Financial Mathematics