1. Chapter 5 Class 11 Complex Numbers
2. Serial order wise

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Misc, 16 If (x + iy)3 = u + iv, then show that u/x + v/y = 4 (π₯2 β π¦2) . We know that (π + π)^3 = π3 + π3 +3ππ (π + π) Replacing a = x and b = iy (π₯ + ππ¦)3= π₯3 + (ππ¦)3 + 3 π₯ ππ¦ (π₯ + ππ¦) = π₯3 + π3π¦3 + 3π₯ π¦π (π₯ + ππ¦) = π₯3 + π2 Γπ π¦3 + 3π₯2π¦π+ 3π₯π¦2π2 Putting π2 = β1 = π₯3 + (β 1 Γ π Γ π₯π¦2) + 3π₯2 π¦π + 3π₯π¦2 π₯(β1) = π₯3 β ππ¦3 + 3π₯2 π¦π β 3π₯π¦2 = π₯3 β 3π₯π¦2 β ππ¦3 + 3π₯2π¦π = π₯3 β 3π₯π¦2 + 3π₯2π¦π β ππ¦3 = π₯3 β 3π₯π¦2 + (3π₯2π¦ β π¦3)π Hence, (π₯ + ππ¦)3 = π₯3 β 3π₯π¦2 + (3π₯2π¦ β π¦3)π But, (π₯ + ππ¦)3 = π’ + ππ£ So, π₯3 β 3π₯π¦2 + (3π₯2π¦ β π¦3)π = π’ + ππ£ Comparing Real parts π₯3 β 3π₯π¦2 = π’ π₯ (π₯2β 3π¦2) = π’ π₯2 β 3π¦2 = π’/π₯ Adding (1) & (2) i.e. (1) + (2) π’/π₯ + π£/π¦ = (π₯2 β 3π¦2) + (3π₯2 βπ¦2) = π₯2 β 3π¦2 +3π₯2 β π¦2 = 4π₯2 β 4π¦2 = 4 (π₯2 β π¦2) Thus, u/x + v/y = 4 (x2 β y2) Hence Proved