2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Complex Numbers and Quadratic Equations

JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Conjugate, Modulus and Argument of complex number

  • question_answer
    If for complex numbers \[{{z}_{1}}\] and \[{{z}_{2}}\], \[arg({{z}_{1}}/{{z}_{2}})=0,\] then \[|{{z}_{1}}-{{z}_{2}}|\] is equal to

    A) \[|{{z}_{1}}|+|{{z}_{2}}|\]

    B) \[|{{z}_{1}}|-|{{z}_{2}}|\]

    C) \[||{{z}_{1}}|-|{{z}_{2}}||\]

    D) 0

    Correct Answer: C

    Solution :

    We have \[|{{z}_{1}}-{{z}_{2}}{{|}^{2}}\] \[=|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}-2|{{z}_{1}}||{{z}_{2}}|\cos ({{\theta }_{1}}-{{\theta }_{2}})\] where \[{{\theta }_{1}}=arg({{z}_{1}})\] and \[{{\theta }_{2}}=arg({{z}_{2}})\] Since \[arg\,{{z}_{1}}-arg\,{{z}_{2}}=0\] \\[|{{z}_{1}}-{{z}_{2}}{{|}^{2}}=|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}-2|{{z}_{1}}||{{z}_{2}}|\]   \[={{(|{{z}_{1}}|-|{{z}_{2}}|)}^{2}}\] Þ \[|{{z}_{1}}-{{z}_{2}}|=||{{z}_{1}}|-|{{z}_{2}}||\]


You need to login to perform this action.
You will be redirected in 3 sec spinner