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Misc 2 - Prove that Re(z1 z2) = Re z1 Re z2 - Im z1 Im z2 - Miscellaneous

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc 2 For any two complex numbers z1 and z2, prove that 𝑅𝑒 (𝑧1𝑧2) = 𝑅𝑒 𝑧1 𝑅𝑒 𝑧2 – πΌπ‘š 𝑧1 πΌπ‘š 𝑧2 Complex number is of form 𝑧 = π‘₯ + 𝑖𝑦 Hence Let complex number 𝑧1 = π‘₯1 + 𝑖𝑦1 Let complex number 𝑧2 = π‘₯2 + 𝑖𝑦2 Solving RHS first 𝑅𝑒 𝑧1 𝑅𝑒 𝑧2 βˆ’ πΌπ‘š 𝑧1 πΌπ‘š 𝑧2 = π‘₯1 π‘₯2 βˆ’π‘¦1 𝑦2 Now, calculating 𝑧1 𝑧2 𝑧1 𝑧2 = ( π‘₯1 + 𝑖𝑦1) (π‘₯2 + 𝑖𝑦2) = π‘₯1 (π‘₯2 + 𝑖𝑦2) + 𝑖𝑦1( π‘₯2 + 𝑖𝑦2) = π‘₯1 π‘₯2 + 𝑖π‘₯1 𝑦2 + π‘₯2 𝑦1 𝑖 +𝑖^2 𝑦1 𝑦2 Putting i2 = βˆ’1 = π‘₯1 π‘₯2 + π‘₯1 𝑦2𝑖 + π‘₯2 𝑦1𝑖 βˆ’ 𝑦1 𝑦2 = π‘₯1 π‘₯2 βˆ’ 𝑦1𝑦2 + (π‘₯1𝑦2 + π‘₯2𝑦1)𝑖 Solving L.H.S 𝑅𝑒 (𝑧1 𝑧2) = Real part of 𝑧1 𝑧2 = π‘₯1 π‘₯2 βˆ’π‘¦1 𝑦2 = R.H.S Hence LHS = RHS Hence Proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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