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Misc 16 - If (x + iy)3 = u + iv, then show that u/x + v/y - Proof- Solving

Misc 16 - Chapter 5 Class 11 Complex Numbers - Part 2
Misc 16 - Chapter 5 Class 11 Complex Numbers - Part 3

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Misc, 16 If (x + iy)3 = u + iv, then show that u/x + v/y = 4 (π‘₯2 – 𝑦2) . We know that (π‘Ž + 𝑏)^3 = π‘Ž3 + 𝑏3 +3π‘Žπ‘ (π‘Ž + 𝑏) Replacing a = x and b = iy (π‘₯ + 𝑖𝑦)3= π‘₯3 + (𝑖𝑦)3 + 3 π‘₯ 𝑖𝑦 (π‘₯ + 𝑖𝑦) = π‘₯3 + 𝑖3𝑦3 + 3π‘₯ 𝑦𝑖 (π‘₯ + 𝑖𝑦) = π‘₯3 + 𝑖2 ×𝑖 𝑦3 + 3π‘₯2𝑦𝑖+ 3π‘₯𝑦2𝑖2 Putting 𝑖2 = –1 = π‘₯3 + (βˆ’ 1 Γ— 𝑖 Γ— π‘₯𝑦2) + 3π‘₯2 𝑦𝑖 + 3π‘₯𝑦2 π‘₯(βˆ’1) = π‘₯3 – 𝑖𝑦3 + 3π‘₯2 𝑦𝑖 βˆ’ 3π‘₯𝑦2 = π‘₯3 – 3π‘₯𝑦2 βˆ’ 𝑖𝑦3 + 3π‘₯2𝑦𝑖 = π‘₯3 – 3π‘₯𝑦2 + 3π‘₯2𝑦𝑖 βˆ’ 𝑖𝑦3 = π‘₯3 – 3π‘₯𝑦2 + (3π‘₯2𝑦 βˆ’ 𝑦3)𝑖 Hence, (π‘₯ + 𝑖𝑦)3 = π‘₯3 – 3π‘₯𝑦2 + (3π‘₯2𝑦 βˆ’ 𝑦3)𝑖 But, (π‘₯ + 𝑖𝑦)3 = 𝑒 + 𝑖𝑣 So, π‘₯3 – 3π‘₯𝑦2 + (3π‘₯2𝑦 βˆ’ 𝑦3)𝑖 = 𝑒 + 𝑖𝑣 Comparing Real parts π‘₯3 – 3π‘₯𝑦2 = 𝑒 π‘₯ (π‘₯2– 3𝑦2) = 𝑒 π‘₯2 – 3𝑦2 = 𝑒/π‘₯ Adding (1) & (2) i.e. (1) + (2) 𝑒/π‘₯ + 𝑣/𝑦 = (π‘₯2 – 3𝑦2) + (3π‘₯2 –𝑦2) = π‘₯2 – 3𝑦2 +3π‘₯2 – 𝑦2 = 4π‘₯2 – 4𝑦2 = 4 (π‘₯2 – 𝑦2) Thus, u/x + v/y = 4 (x2 – y2) Hence Proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.