Q52.Let z be a complex number such that z+2i z−i = 1 and |z| = 52 . Then, the value of |z + 3i| is (1) √10 (2) 72 (3) 15 (4) 2√3 4
What This Question Tests
This question combines the algebraic and geometric interpretations of complex numbers, requiring the use of modulus properties to find a complex number and then compute the modulus of another related complex number.
Concepts Tested
Formulas Used
|z-a| = |z-b| implies Re(z) = Re((a+b)/2)
|z|^2 = x^2 + y^2
📚 NCERT Sections This Tests
5.29 — Amongst The Following Ions Which One Has The Highest Magnetic Moment Value?
Chemistry Class 11 · Chapter 5
5.29 Amongst the following ions which one has the highest magnetic moment value? (i) [Cr(H2O)6]3+ (ii) [Fe(H2O)6] 2+ (iii) [Zn(H2O)6]2+ 5.30 Amongst the following, the most stable complex is (i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6] 3+ (iii) [Fe(C2O4)3]3– (iv) [FeCl6] 3– 5.31 What will be the correct order for the wavelengths of absorption in the visible region for the following: [Ni(NO2)6] 4–, [Ni(NH3)6] 2+, [Ni(H2O)6] 2+ ? Answers to Some Intext Questions 5.1 (i) [Co(NH3)4(H2O)2]Cl3 (iv) [Pt(NH3)BrCl(NO2)]– (ii) K2[Ni(CN)4] (v) [PtCl2(en)2](NO3)2 (iii) [Cr(en)3]Cl3 (vi) Fe4[Fe(CN)6]3 5.2 (i) Hexaamminecobalt(III) chloride (ii) Pentaamminechloridocobalt(III) chloride (iii) Potassium hexacyanidoferrate(III) (iv) Potassium trioxalatoferrate(III) (v) Potassium tetrachloridopalladate(II) (vi) Diamminechlorido(methanamine)platinum(II) chloride 5.3 (i) Both geometrical (cis-, trans-) and optical isomers for cis can exist. (ii) Two optical isomers can exist. (iii) There are 10 possible isomers. (Hint: There are geometrical, ionisation and linkage isomers possible). (iv) Geometrical (cis-, trans-) isomers can exist. 5.4 The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents: [Co(NH3)5Br]SO4 + Ba2+ ® BaSO4 (s) [Co(NH3)5SO4]Br + Ba2+ ® No reaction [Co(NH3)5Br]SO4 + Ag+ ® No reaction [Co(NH3)5SO4]Br + Ag+ ® AgBr (s) 5.6 In Ni(CO)4, Ni is in zero oxidation state whereas in NiCl42–, it is in +2 oxidation state. In the presence of CO ligand, the unpaired d electrons of Ni pair up but Cl– being a weak ligand is unable to pair up the unpaired electrons. 5.7 In presence of CN–, (a strong ligand) the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d 2sp 3 forming inner orbital complex. In the presence of H2O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp 3d 2 forming an outer orbital complex containing five unpaired electrons, it is strongly paramagnetic. 5.8 In the presence of NH3, the 3d electrons pair up leaving two d orbitals empty to be involved in d2sp3 hybridisation forming inner orbital complex in case of [Co(NH3)6]3+. In Ni(NH3)6 2+, Ni is in +2 oxidation state and has d 8 configuration, the hybridisation involved is sp 3d 2 forming outer orbital complex. 5.9 For square planar shape, the hybridisation is dsp 2. Hence the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp2 hybridisation. Thus there is no unpaired electron. Chemistry 140 Reprint 2025-26
5.11 — Draw All The Isomers (Geometrical And Optical) Of:
Chemistry Class 11 · Chapter 5
5.11 Draw all the isomers (geometrical and optical) of: (i) [CoCl2(en)2] + (ii) [Co(NH3)Cl(en)2] 2+ (iii) [Co(NH3)2Cl2(en)]+
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
📋 Question Details
- Chapter
- Complex Numbers
- Topic
- Modulus of complex numbers, Locus of complex numbers
- Year
- 2020
- Shift
- 09 Jan Shift 1
- Q Number
- Q52
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 5: Complex Numbers and Quadratic Equations
More from this Chapter
Q84.If |z + 4| ≤3 , then the maximum value of |z + 1| is (1) 4 (2) 10 (3) 6 (4) 0
Q73.The conjugate of a complex number is 1 . Then the complex number is i−1 (1) −1 (2) 1 i−1 i+1 (3) −1 (4) 1 i+1 i−1
Q62.If z −4z = 2, then the maximum value of |z| is equal to (1) √3 + 1 (2) √5 + 1 (3) 2 (4) 2 + √2
Q61.If α and β are the roots of the equation x2 −x + 1 = 0, then α2009 + β2009 = (1) −1 (2) 1 (3) 2 (4) −2