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

  1. Chapter 5 Class 11 Complex Numbers
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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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