Complex Numbers – Exercise 2.9 Answer Key (Class 12 Maths)

Quick-reference answers for Exercise 2.9 (choose the correct option). Each line gives the correct option number and a one-line justification; full worked solutions appear with each question in the practice set.

1. Q1. Option (1) — A. Four consecutive powers of \(i\) always sum to zero.
2. Q2. Option (1) — A. Factor \(i^{n-1}(1+i)\); the geometric sum collapses to \(1+i\).
3. Q3. Option (1) — A. \(z,iz,z+iz\) are three corners of a square of area \(|z|^{2}\); the triangle is half of it.
4. Q4. Option (2) — B. Conjugating \(\frac{1}{i-2}\) gives \(\frac{1}{-i-2}=\frac{-1}{i+2}\).
5. Q5. Option (3) — C. \(|z|=\frac{2^{3}\cdot 5^{2}}{10^{2}}=2\) by multiplicativity of modulus.
6. Q6. Option (1) — A. Taking modulus: \(2|z|^{2}=|z|\Rightarrow|z|=\frac12\).
7. Q7. Option (4) — D. Disc centred at \(2-i\), radius 2; max \(|z|=|2-i|+2=\sqrt5+2\).
8. Q8. Option (1) — A. Reverse triangle inequality gives \(r^{2}+2r-3\ge0\Rightarrow r\ge1\).
9. Q9. Option (1) — A. \(\bar z=\frac1z\) when \(|z|=1\), so the ratio simplifies to \(z\).
10. Q10. Option (1) — A. \(y=-2\) and \(\sqrt{x^{2}+4}=x+1\Rightarrow x=\frac32\).
11. Q11. Option (2) — B. Factor \(z_{1}z_{2}z_{3}\); modulus gives \(6|z_1+z_2+z_3|=12\).
12. Q12. Option (2) — B. \(z+\frac1z\) real with \(z\) non-real forces \(|z|^{2}=1\).
13. Q13. Option (4) — D. \((\sum z)^{2}-2\sum z_iz_j=0-0=0\).
14. Q14. Option (2) — B. Purely imaginary \(\Rightarrow w+\bar w=0\Rightarrow|z|^{2}=1\).
15. Q15. Option (2) — B. Equidistant from \(\pm2\); locus is the line \(x=0\).
16. Q16. Option (3) — C. \(\cos50^\circ+i\sin50^\circ\) to the 5th gives arg \(250^\circ\equiv-110^\circ\).
17. Q17. Option (1) — A. Third-quadrant point; principal argument \(-(\pi-\frac\pi4)=-\frac{3\pi}{4}\).
18. Q18. Option (3) — C. \(\frac{1+i}{1-i}=i\), so \(i^{n}=1\) first holds at \(n=4\).
19. Q19. Option (4) — D. \((1+\omega)^{7}=-\omega^{2}=1+\omega\Rightarrow(A,B)=(1,1)\).
20. Q20. Option (4) — D. Simplifies to \(\frac12\operatorname{cis}\frac{\pi}{2}\); principal argument \(=\frac{\pi}{2}\).
21. Q21. Option (2) — B. Roots \(\cos60^\circ\pm i\sin60^\circ\); \(2\cos(2020\cdot60^\circ)=-1\).
22. Q22. Option (3) — C. Four fourth-roots of \(-1\); product \(=(-1)^{5}(-1)=1\).
23. Q23. Option (4) — D. Determinant \(=3(\omega^{2}-\omega)=-3\sqrt3\,i\).
24. Q24. Option (1) — A. \(i^{4n+1}=i,\;i^{4n-1}=-i\); \(\frac{i+i}{2}=i\).
25. Q25. Option (1) — A. Determinant simplifies to \(z^{3}\); \(z^{3}=0\) has one distinct root.