Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    The vector \[z=-4+5i\]is turned counter clockwise through an angle of\[{{180}^{o}}\]and stretched\[1\frac{1}{2}\]times. The complex number corresponding to newly obtained vector is

    A) \[-6+\frac{15}{2}i\]         

    B)        \[6+\frac{15}{2}i\]

    C) \[6-\frac{15}{2}i\]           

    D)         None of these

    Correct Answer: C

    Solution :

    We know that, when we multiply a complex number by\[i\], it rotates the vector for\[z\]by an angle of\[\frac{\pi }{2}\]in the anticlockwise direction. \[\therefore \]Multiplying by \[{{i}^{2}}\]will rotate the vector for\[z\]by\[\pi \]. \[\therefore \]  \[z={{i}^{2}}(-4+5i)=-(-4+5i)\] Stretching\[\frac{3}{2}\]times \[\therefore \]  \[z\,\,=\frac{3}{2}(4-5i)\]                 \[=6-\frac{15}{2}i\]


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