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Misc 10 - Chapter 4 Class 11 Complex Numbers
Last updated at July 26, 2026 by Teachoo
Chapter 4 Class 11 Complex Numbers
Concept wise
- Quadaratic equation
- Two complex numbers equal
- Convert to a + ib form
- Identities (square, cube of 2 complex numbers)
- Division
- Power of i(odd and even)
- Conjugate
- Modulus,argument
- Polar representation
- Proof- Replacing iota with -iota
- Proof- Taking general complex numbers
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Proof- Solving
- Proof- Using modulus conjugate property (Pg 102)
Transcript
Misc, 16 If (x + iy)3 = u + iv, then show that u/x + v/y = 4 (š„2 ā š¦2) . We know that (š + š)^3 = š3 + š3 +3šš (š + š) Replacing a = x and b = iy (š„ + šš¦)3= š„3 + (šš¦)3 + 3 š„ šš¦ (š„ + šš¦) = š„3 + š3š¦3 + 3š„ š¦š (š„ + šš¦) = š„3 + š2 Ćš š¦3 + 3š„2š¦š+ 3š„š¦2š2 Putting š2 = ā1 = š„3 + (ā 1 Ć š Ć š„š¦2) + 3š„2 š¦š + 3š„š¦2 š„(ā1) = š„3 ā šš¦3 + 3š„2 š¦š ā 3š„š¦2 = š„3 ā 3š„š¦2 ā šš¦3 + 3š„2š¦š = š„3 ā 3š„š¦2 + 3š„2š¦š ā šš¦3 = š„3 ā 3š„š¦2 + (3š„2š¦ ā š¦3)š Hence, (š„ + šš¦)3 = š„3 ā 3š„š¦2 + (3š„2š¦ ā š¦3)š But, (š„ + šš¦)3 = š¢ + šš£ So, š„3 ā 3š„š¦2 + (3š„2š¦ ā š¦3)š = š¢ + šš£ Comparing Real parts š„3 ā 3š„š¦2 = š¢ š„ (š„2ā 3š¦2) = š¢ š„2 ā 3š¦2 = š¢/š„ Adding (1) & (2) i.e. (1) + (2) š¢/š„ + š£/š¦ = (š„2 ā 3š¦2) + (3š„2 āš¦2) = š„2 ā 3š¦2 +3š„2 ā š¦2 = 4š„2 ā 4š¦2 = 4 (š„2 ā š¦2) Thus, u/x + v/y = 4 (x2 ā y2) Hence Proved
