11th Class Mathematics Complex Numbers and Quadratic Equations Question Bank Critical Thinking

  • question_answer If \[{{z}_{1}}=a+ib\] and \[{{z}_{2}}=c+id\] are complex numbers such that  \[|{{z}_{1}}|\,=\,|{{z}_{2}}|=1\] and \[R({{z}_{1}}\overline{{{z}_{2}}})=0,\] then the pair of complex numbers \[{{w}_{1}}=a+ic\] and \[{{w}_{2}}=b+id\] satisfies [IIT 1985]

    A) \[|{{w}_{1}}|=1\]

    B) \[|{{w}_{2}}|=1\]

    C) \[R({{w}_{1}}\overline{{{w}_{2}}})=0,\]

    D) All the above

    Correct Answer: D

    Solution :

    Since\[|{{z}_{1}}|\,=\,|{{z}_{2}}|\,=1\], we have \[{{z}_{1}}=\cos {{\theta }_{1}}+i\sin {{\theta }_{1}},{{z}_{2}}=\cos {{\theta }_{2}}+i\sin {{\theta }_{2}}\] where \[{{\theta }_{1}}=arg({{z}_{1}})\] and \[{{\theta }_{2}}=arg({{z}_{2}})\] Also, \[{{z}_{1}}=a+ib\] and \[{{z}_{2}}=c+id.\] Therefore\[a=\cos {{\theta }_{1}}\],\[b=\sin {{\theta }_{1}},c=\cos {{\theta }_{2}}\] and \[d=\sin {{\theta }_{2}}\] Also, \[R({{z}_{1}}{{\overline{z}}_{2}})=0\] Þ \[R[(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})(\cos {{\theta }_{2}}-i\sin {{\theta }_{2}})]=0\] Þ \[R[(\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})]=0\] Þ \[\cos ({{\theta }_{1}}-{{\theta }_{2}})=0\]Þ\[{{\theta }_{1}}-{{\theta }_{2}}=\frac{\pi }{2}\]Þ \[{{\theta }_{1}}={{\theta }_{2}}+\frac{\pi }{2}\] Now, \[{{w}_{1}}=a+ic=\cos {{\theta }_{1}}+i\cos {{\theta }_{2}}\]\[=\cos {{\theta }_{1}}+i\sin {{\theta }_{1}}\] Þ \[|{{w}_{1}}|\,=1\] Similarly, \[|{{w}_{2}}|\,=1\] Next \[{{w}_{1}}{{\overline{w}}_{2}}=(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})(\cos {{\theta }_{2}}-i\sin {{\theta }_{2}})\]                   \[=\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})\]Þ \[|{{w}_{1}}{{\overline{w}}_{2}}|\,=1\] Finally, \[R({{\overline{w}}_{1}}{{w}_{2}})=R({{w}_{2}}{{\overline{w}}_{1}})\] \[=R[(\cos {{\theta }_{2}}+i\sin {{\theta }_{2}})(\cos {{\theta }_{1}}-i\sin {{\theta }_{1}})]\] \[=R[\cos ({{\theta }_{2}}-{{\theta }_{1}})+i\sin ({{\theta }_{2}}-{{\theta }_{1}})]\] \[=\cos ({{\theta }_{2}}-{{\theta }_{1}})=\cos \left( \frac{-\pi }{2} \right)=0\]


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