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

  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 (๐‘ฅ2 โ€“ ๐‘ฆ2) . We know that (๐‘Ž + ๐‘)^3 = ๐‘Ž3 + ๐‘3 +3๐‘Ž๐‘ (๐‘Ž + ๐‘) Replacing a = x and b = iy (๐‘ฅ + ๐‘–๐‘ฆ)3= ๐‘ฅ3 + (๐‘–๐‘ฆ)3 + 3 ๐‘ฅ ๐‘–๐‘ฆ (๐‘ฅ + ๐‘–๐‘ฆ) = ๐‘ฅ3 + ๐‘–3๐‘ฆ3 + 3๐‘ฅ ๐‘ฆ๐‘– (๐‘ฅ + ๐‘–๐‘ฆ) = ๐‘ฅ3 + ๐‘–2 ร—๐‘– ๐‘ฆ3 + 3๐‘ฅ2๐‘ฆ๐‘–+ 3๐‘ฅ๐‘ฆ2๐‘–2 Putting ๐‘–2 = โ€“1 = ๐‘ฅ3 + (โˆ’ 1 ร— ๐‘– ร— ๐‘ฅ๐‘ฆ2) + 3๐‘ฅ2 ๐‘ฆ๐‘– + 3๐‘ฅ๐‘ฆ2 ๐‘ฅ(โˆ’1) = ๐‘ฅ3 โ€“ ๐‘–๐‘ฆ3 + 3๐‘ฅ2 ๐‘ฆ๐‘– โˆ’ 3๐‘ฅ๐‘ฆ2 = ๐‘ฅ3 โ€“ 3๐‘ฅ๐‘ฆ2 โˆ’ ๐‘–๐‘ฆ3 + 3๐‘ฅ2๐‘ฆ๐‘– = ๐‘ฅ3 โ€“ 3๐‘ฅ๐‘ฆ2 + 3๐‘ฅ2๐‘ฆ๐‘– โˆ’ ๐‘–๐‘ฆ3 = ๐‘ฅ3 โ€“ 3๐‘ฅ๐‘ฆ2 + (3๐‘ฅ2๐‘ฆ โˆ’ ๐‘ฆ3)๐‘– Hence, (๐‘ฅ + ๐‘–๐‘ฆ)3 = ๐‘ฅ3 โ€“ 3๐‘ฅ๐‘ฆ2 + (3๐‘ฅ2๐‘ฆ โˆ’ ๐‘ฆ3)๐‘– But, (๐‘ฅ + ๐‘–๐‘ฆ)3 = ๐‘ข + ๐‘–๐‘ฃ So, ๐‘ฅ3 โ€“ 3๐‘ฅ๐‘ฆ2 + (3๐‘ฅ2๐‘ฆ โˆ’ ๐‘ฆ3)๐‘– = ๐‘ข + ๐‘–๐‘ฃ Comparing Real parts ๐‘ฅ3 โ€“ 3๐‘ฅ๐‘ฆ2 = ๐‘ข ๐‘ฅ (๐‘ฅ2โ€“ 3๐‘ฆ2) = ๐‘ข ๐‘ฅ2 โ€“ 3๐‘ฆ2 = ๐‘ข/๐‘ฅ Adding (1) & (2) i.e. (1) + (2) ๐‘ข/๐‘ฅ + ๐‘ฃ/๐‘ฆ = (๐‘ฅ2 โ€“ 3๐‘ฆ2) + (3๐‘ฅ2 โ€“๐‘ฆ2) = ๐‘ฅ2 โ€“ 3๐‘ฆ2 +3๐‘ฅ2 โ€“ ๐‘ฆ2 = 4๐‘ฅ2 โ€“ 4๐‘ฆ2 = 4 (๐‘ฅ2 โ€“ ๐‘ฆ2) Thus, u/x + v/y = 4 (x2 โ€“ y2) Hence Proved

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