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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If\[z=\sqrt{2}-i\sqrt{2}\]is rotated through an angle\[45{}^\circ \]in the anticlockwise direction about the origin, then the co-ordinates of its new position are:

    A) \[(2,0)\]              

    B)         \[(\sqrt{2},\sqrt{2})\]

    C)  \[(\sqrt{2},-\sqrt{2})\] 

    D)         \[(\sqrt{2},0)\]

    E)  \[(4,0)\]

    Correct Answer: A

    Solution :

    \[{{z}_{2}}=z{{e}^{i\pi /4}}=(\sqrt{2}-i\sqrt{2})(\cos \pi /4+i\sin \pi /4)\] \[=\sqrt{2}(1-i)\frac{(1+i)}{\sqrt{2}}\]                 \[=1-{{i}^{2}}=1+1=2+0i\] \[\therefore \]Co-ordinate of new position are (2, 0).


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