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Level 2 — RecallComplex Numbers

Complex Numbers

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level: 2 (Recall & standard textbook problems) Time limit: 30 minutes Total marks: 40

Answer all questions. Use i2=−1i^2=-1 throughout. Show working where relevant.

Q1. Simplify the following powers of ii: (a) i15i^{15}, (b) i100i^{100}, (c) i−7i^{-7}. (3 marks)

Q2. For z=3−4iz = 3 - 4i, write down: (a) the real part, (b) the imaginary part, (c) the complex conjugate zˉ\bar z, (d) the modulus ∣z∣|z|. (4 marks)

Q3. Express z=−1+3 iz = -1 + \sqrt{3}\,i in polar form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta), giving the argument in radians. (4 marks)

Q4. Evaluate and write the result in the form a+bia+bi: 2+3i1−2i.\frac{2+3i}{1-2i}. (4 marks)

Q5. If z1=2 cis 40∘z_1 = 2\,\mathrm{cis}\,40^\circ and z2=3 cis 20∘z_2 = 3\,\mathrm{cis}\,20^\circ, find z1z2z_1 z_2 and z1z2\dfrac{z_1}{z_2} in polar (cis) form. (4 marks)

Q6. State De Moivre's theorem. Hence evaluate (1+i)8(1+i)^8, giving your answer as a real/complex number in the form a+bia+bi. (5 marks)

Q7. State Euler's formula and use it to express eiπe^{i\pi} and eiπ/2e^{i\pi/2} as simplified complex numbers. (4 marks)

Q8. Find all three cube roots of unity, and show that their sum is zero. (5 marks)

Q9. Find all the values of zz satisfying z2=5−12iz^2 = 5 - 12i. (4 marks)

Q10. Solve the quadratic equation z2−4z+13=0z^2 - 4z + 13 = 0, expressing the roots in the form a+bia+bi. Verify they are complex conjugates. (3 marks)

End of paper

Answer keyMark scheme & solutions

Q1. Powers of ii cycle with period 4 (reduce exponent mod 4).

  • (a) 15≡3(mod4)⇒i15=i3=−i15 \equiv 3 \pmod 4 \Rightarrow i^{15}=i^3=-i. (1)
  • (b) 100≡0(mod4)⇒i100=1100 \equiv 0 \pmod 4 \Rightarrow i^{100}=1. (1)
  • (c) i−7=i−7+8=i1=ii^{-7}=i^{-7+8}=i^{1}=i. (1)

Why: i4=1i^4=1, so only the remainder of the exponent mod 4 matters; negatives are shifted by adding multiples of 4.

Q2. For z=3−4iz=3-4i:

  • (a) Re⁡(z)=3\operatorname{Re}(z)=3 (1)
  • (b) Im⁡(z)=−4\operatorname{Im}(z)=-4 (1)
  • (c) zˉ=3+4i\bar z = 3+4i (1)
  • (d) ∣z∣=32+(−4)2=25=5|z|=\sqrt{3^2+(-4)^2}=\sqrt{25}=5 (1)

Why: conjugate flips sign of imaginary part; modulus is distance from origin.

Q3. z=−1+3 iz=-1+\sqrt3\,i.

  • r=(−1)2+(3)2=1+3=2r=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2. (1)
  • Point in 2nd quadrant. Reference angle tan⁡−1(3/1)=60∘=π/3\tan^{-1}(\sqrt3/1)=60^\circ=\pi/3; so θ=π−π/3=2π3\theta=\pi-\pi/3=\tfrac{2\pi}{3}. (2)
  • z=2(cos⁡2π3+isin⁡2π3)z=2\left(\cos\tfrac{2\pi}{3}+i\sin\tfrac{2\pi}{3}\right). (1)

Why: argument must respect the quadrant of the point.

Q4. Multiply by conjugate of denominator 1+2i1+2i: (2+3i)(1+2i)(1−2i)(1+2i)=2+4i+3i+6i21+4=2+7i−65=−4+7i5.\frac{(2+3i)(1+2i)}{(1-2i)(1+2i)}=\frac{2+4i+3i+6i^2}{1+4}=\frac{2+7i-6}{5}=\frac{-4+7i}{5}. Working (2), simplification (1), result =−45+75i=-\tfrac{4}{5}+\tfrac{7}{5}i. (1)

Q5. Multiply: moduli multiply, arguments add. z1z2=(2⋅3) cis(40∘+20∘)=6 cis 60∘.z_1z_2 = (2\cdot3)\,\mathrm{cis}(40^\circ+20^\circ)=6\,\mathrm{cis}\,60^\circ. (2) Divide: moduli divide, arguments subtract. z1z2=23 cis(40∘−20∘)=23 cis 20∘.\frac{z_1}{z_2}=\frac{2}{3}\,\mathrm{cis}(40^\circ-20^\circ)=\frac{2}{3}\,\mathrm{cis}\,20^\circ. (2)

Q6. Statement: For integer nn, (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta. (1)

1+i1+i: r=2r=\sqrt2, θ=45∘=π/4\theta=45^\circ=\pi/4. (1) (1+i)8=(2)8 cis(8⋅45∘)=24 cis 360∘=16(cos⁡360∘+isin⁡360∘).(1+i)^8=(\sqrt2)^8\,\mathrm{cis}(8\cdot45^\circ)=2^4\,\mathrm{cis}\,360^\circ=16(\cos360^\circ+i\sin360^\circ). (2) =16(1+0i)=16=16(1+0i)=16. (1)

Q7. Euler's formula: eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta. (2)

  • eiπ=cos⁡π+isin⁡π=−1+0i=−1e^{i\pi}=\cos\pi+i\sin\pi=-1+0i=-1. (1)
  • eiπ/2=cos⁡π2+isin⁡π2=0+i=ie^{i\pi/2}=\cos\tfrac\pi2+i\sin\tfrac\pi2=0+i=i. (1)

Q8. Cube roots of unity: solve z3=1z^3=1, i.e. z=cis(2πk3)z=\mathrm{cis}\big(\tfrac{2\pi k}{3}\big), k=0,1,2k=0,1,2. (1)

  • k=0: z=1k=0:\ z=1
  • k=1: z=cos⁡120∘+isin⁡120∘=−12+32i (=ω)k=1:\ z=\cos120^\circ+i\sin120^\circ=-\tfrac12+\tfrac{\sqrt3}{2}i \ (=\omega)
  • k=2: z=cos⁡240∘+isin⁡240∘=−12−32i (=ω2)k=2:\ z=\cos240^\circ+i\sin240^\circ=-\tfrac12-\tfrac{\sqrt3}{2}i \ (=\omega^2) (3)

Sum: 1+(−12+32i)+(−12−32i)=01+\left(-\tfrac12+\tfrac{\sqrt3}{2}i\right)+\left(-\tfrac12-\tfrac{\sqrt3}{2}i\right)=0. (1)

Why: the roots are equally spaced on the unit circle; their vector sum cancels.

Q9. Let z=x+iyz=x+iy with z2=5−12iz^2=5-12i. x2−y2=5,2xy=−12⇒xy=−6.x^2-y^2=5,\qquad 2xy=-12\Rightarrow xy=-6. (1) Also ∣z∣2=∣w∣1/...|z|^2=|w|^{1/... }: x2+y2=∣5−12i∣=25+144=13x^2+y^2=|5-12i|=\sqrt{25+144}=13. (1) Adding: 2x2=18⇒x2=9⇒x=±32x^2=18\Rightarrow x^2=9\Rightarrow x=\pm3; then y=−6/xy=-6/x, so x=3⇒y=−2x=3\Rightarrow y=-2; x=−3⇒y=2x=-3\Rightarrow y=2. (1) Roots: z=3−2iz=3-2i or z=−3+2iz=-3+2i. (1)

Check: (3−2i)2=9−12i+4i2=5−12i.(3-2i)^2=9-12i+4i^2=5-12i. ✓

Q10. z2−4z+13=0z^2-4z+13=0. Discriminant =16−52=−36=16-52=-36. z=4±−362=4±6i2=2±3i.z=\frac{4\pm\sqrt{-36}}{2}=\frac{4\pm6i}{2}=2\pm3i. (2) Roots 2+3i2+3i and 2−3i2-3i are conjugates (real coefficients guarantee this). (1)

[
  {"claim":"i^15=-i, i^100=1, i^-7=i","code":"result = (I**15==-I) and (I**100==1) and (I**-7==I)"},
  {"claim":"(2+3i)/(1-2i)=-4/5+7/5 i","code":"result = simplify((2+3*I)/(1-2*I) - (Rational(-4,5)+Rational(7,5)*I))==0"},
  {"claim":"(1+i)^8=16","code":"result = expand((1+I)**8)==16"},
  {"claim":"cube roots of unity sum to 0","code":"w=Rational(-1,2)+sqrt(3)/2*I; result = simplify(1+w+w**2)==0"},
  {"claim":"z^2=5-12i has root 3-2i","code":"result = expand((3-2*I)**2)==5-12*I"},
  {"claim":"z^2-4z+13=0 roots 2+/-3i","code":"result = set(solve(z**2-4*z+13, z))=={2+3*I,2-3*I}"}
]