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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    Let a =Im\[\left( \frac{1+{{z}^{2}}}{2iz} \right),\]where z is any non- zero complex ?complex number.                 The set \[\operatorname{A}=\{a:\left| z \right|1\operatorname{and}z\ne \pm 1\}\] is equal to :     JEE Main Online Paper ( Held On 23  April 2013 )

    A)                   (-1,1)                  

    B)                                          [-1,1]

    C)                                          [0, 1)                   

    D)                                          (-1,0]

    Correct Answer: A

    Solution :

                      Let\[z=x+iy\Rightarrow {{z}^{2}}={{x}^{2}}-{{y}^{2}}+2ixy\]Now, \[\frac{1+{{z}^{2}}}{2iz}=\frac{1+{{x}^{2}}-{{y}^{2}}}{2i(x+iy)}=\frac{({{x}^{2}}-{{y}^{2}}+1)+2ixy}{2ix-2y}\] \[\frac{({{x}^{2}}-{{y}^{2}}+1)+2ixy}{-2y+2ix}\times \frac{-2y-2ix}{-2y-2ix}\] \[=\frac{y({{x}^{2}}-{{y}^{2}}-1)+x({{x}^{2}}+{{y}^{2}}+1)i}{2({{x}^{2}}+{{y}^{2}})}\] \[a=\frac{x({{x}^{2}}-{{y}^{2}}+1)}{2({{x}^{2}}+{{y}^{2}})}\] Since, \[|z|=1\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=1\]\[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}=1\] \[\therefore \]\[a=\frac{x(1+1)}{2\times 1}=x\] Also\[z\ne 1\Rightarrow x+iy\ne 1\] \[\therefore \]\[A=(-1,1)\]


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