Level 1 — RecognitionComplex Numbers

Complex Numbers

20 minutes30 marksprintable — key stays hidden on paper

Chapter: 3.5 Complex Numbers Level: 1 — Recognition (MCQ, Matching, True/False with justification) Time limit: 20 minutes Total marks: 30


Section A — Multiple Choice (1 mark each) — 10 marks

Q1. The value of i15i^{15} is: (a) 11 (b) 1-1 (c) ii (d) i-i

Q2. The real and imaginary parts of z=47iz = 4 - 7i are: (a) 4, 74,\ 7 (b) 4, 74,\ -7 (c) 7, 4-7,\ 4 (d) 7, 47,\ 4

Q3. The modulus of z=3+4iz = 3 + 4i is: (a) 55 (b) 77 (c) 7\sqrt{7} (d) 2525

Q4. The complex conjugate of z=2+5iz = -2 + 5i is: (a) 25i2 - 5i (b) 25i-2 - 5i (c) 2+5i2 + 5i (d) 5+2i-5 + 2i

Q5. Euler's formula states that eiθe^{i\theta} equals: (a) cosθisinθ\cos\theta - i\sin\theta (b) sinθ+icosθ\sin\theta + i\cos\theta (c) cosθ+isinθ\cos\theta + i\sin\theta (d) r(cosθ+isinθ)r(\cos\theta + i\sin\theta)

Q6. In the Argand plane, the number z=03iz = 0 - 3i lies on: (a) the positive real axis (b) the negative imaginary axis (c) the positive imaginary axis (d) the origin

Q7. De Moivre's theorem gives (cosθ+isinθ)n(\cos\theta + i\sin\theta)^n equal to: (a) cos(nθ)+isin(nθ)\cos(n\theta) + i\sin(n\theta) (b) ncosθ+insinθn\cos\theta + i\,n\sin\theta (c) cosθn+isinθn\cos\theta^n + i\sin\theta^n (d) cos(nθ)isin(nθ)\cos(n\theta) - i\sin(n\theta)

Q8. The number of distinct nnth roots of a non-zero complex number is: (a) 11 (b) 22 (c) n1n-1 (d) nn

Q9. The product (2+3i)(23i)(2 + 3i)(2 - 3i) equals: (a) 49i4 - 9i (b) 1313 (c) 5-5 (d) 4+9i4 + 9i

Q10. The polar (exponential) form of a complex number with modulus rr and argument θ\theta is: (a) r+eiθr + e^{i\theta} (b) reiθr\,e^{i\theta} (c) erθe^{r\theta} (d) reθr\,e^{\theta}


Section B — Matching (1 mark each) — 5 marks

Q11. Match each expression in Column X with its value in Column Y.

Column X Column Y
(i) i2i^2 (P) ii
(ii) i3i^3 (Q) 11
(iii) i4i^4 (R) 1-1
(iv) i21i^{21} (S) i-i
(v) 1i\dfrac{1}{i} (T) i-i

Write the matches, e.g. (i)–(?).


Section C — True/False WITH justification (2 marks each: 1 verdict + 1 reason) — 15 marks

Q12. The sum of a complex number and its conjugate is always real.

Q13. z1z2=z1+z2|z_1 z_2| = |z_1| + |z_2| for all complex numbers.

Q14. The three cube roots of unity are equally spaced on the unit circle at angles 120°120° apart.

Q15. The equation x2+1=0x^2 + 1 = 0 has no solutions in the complex number system.

Q16. For z=1+iz = 1 + i, the argument is arg(z)=π4\arg(z) = \dfrac{\pi}{4}.

Q17. Dividing complex numbers is carried out by multiplying numerator and denominator by the conjugate of the denominator.

Q18. The sum of all nnth roots of unity is equal to 11.


Answer keyMark scheme & solutions

Section A (1 mark each)

Q1 — (d) i-i. Powers of ii cycle with period 4. 15=4(3)+315 = 4(3)+3, so i15=i3=ii^{15}=i^{3}=-i. (Reduce exponent mod 4.)

Q2 — (b) 4, 74,\ -7. For a+bia+bi, real part a=4a=4, imaginary part b=7b=-7 (the coefficient of ii, including its sign).

Q3 — (a) 55. 3+4i=32+42=25=5|3+4i|=\sqrt{3^2+4^2}=\sqrt{25}=5.

Q4 — (b) 25i-2 - 5i. Conjugate flips the sign of the imaginary part only.

Q5 — (c) cosθ+isinθ\cos\theta + i\sin\theta. Direct statement of Euler's formula.

Q6 — (b) negative imaginary axis. Real part =0=0, imaginary part =3<0=-3<0, so it sits on the imaginary axis below the origin.

Q7 — (a) cos(nθ)+isin(nθ)\cos(n\theta)+i\sin(n\theta). Statement of De Moivre's theorem.

Q8 — (d) nn. A non-zero complex number has exactly nn distinct nnth roots.

Q9 — (b) 1313. (2+3i)(23i)=22+32=4+9=13(2+3i)(2-3i)=2^2+3^2=4+9=13 (product of conjugates =z2=|z|^2).

Q10 — (b) reiθr\,e^{i\theta}. Exponential form of a complex number.

Section B

Q11 (1 mark each):

  • (i) i2=1i^2 = -1(R)
  • (ii) i3=ii^3 = -i(S) (or T, both give i-i; accept either)
  • (iii) i4=1i^4 = 1(Q)
  • (iv) i21=i21mod4=i1=ii^{21}=i^{21\bmod 4}=i^1=i(P)
  • (v) 1i=1iii=i1=i\dfrac{1}{i}=\dfrac{1}{i}\cdot\dfrac{-i}{-i}=\dfrac{-i}{1}=-i(T) (or S)

Award full 5 marks if all values correct; accept the S/T ambiguity for i-i.

Section C (1 mark verdict + 1 mark justification)

Q12 — TRUE. If z=a+biz=a+bi, then z+zˉ=(a+bi)+(abi)=2az+\bar z = (a+bi)+(a-bi)=2a, which is real (imaginary parts cancel).

Q13 — FALSE. The correct modulus rule is z1z2=z1z2|z_1 z_2| = |z_1|\,|z_2| (product, not sum). Counter-example: z1=z2=iz_1=z_2=i gives z1z2=1=1|z_1z_2|=|-1|=1 but z1+z2=2|z_1|+|z_2|=2.

Q14 — TRUE. Cube roots of unity are ei0,ei2π/3,ei4π/3e^{i\cdot 0},\,e^{i2\pi/3},\,e^{i4\pi/3}, i.e. at 0°,120°,240°0°,120°,240° — equally spaced 120°120° apart on the unit circle.

Q15 — FALSE. x2+1=0x2=1x=±ix^2+1=0\Rightarrow x^2=-1\Rightarrow x=\pm i; complex numbers were introduced precisely to solve this. Two solutions ±i\pm i exist.

Q16 — TRUE. z=1+iz=1+i lies in the first quadrant; arg(z)=arctan(1/1)=π/4\arg(z)=\arctan(1/1)=\pi/4.

Q17 — TRUE. Multiplying by the conjugate of the denominator makes the denominator real (zzˉ=z2z\bar z=|z|^2), enabling division into a+bia+bi form.

Q18 — FALSE. The sum of all nnth roots of unity is 00 (for n2n\ge 2), since they are roots of xn1=0x^n-1=0 whose coefficient of xn1x^{n-1} is 00. (Only for n=1n=1 is the sum 11.)

[
  {"claim":"i^15 = -i","code":"result = (I**15 == -I)"},
  {"claim":"|3+4i| = 5","code":"result = (Abs(3+4*I) == 5)"},
  {"claim":"(2+3i)(2-3i) = 13","code":"result = (expand((2+3*I)*(2-3*I)) == 13)"},
  {"claim":"sum of cube roots of unity is 0","code":"roots=[exp(2*pi*I*k/3) for k in range(3)]; result = (simplify(sum(roots))==0)"},
  {"claim":"1/i = -i","code":"result = (simplify(1/I) == -I)"},
  {"claim":"arg(1+i) = pi/4","code":"result = (arg(1+I) == pi/4)"}
]