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

  • question_answer If \[i=\sqrt{-1},\] then \[4+5{{\left( -\frac{1}{2}+\frac{i\sqrt{3}}{2} \right)}^{334}}\]   \[+3{{\left( -\frac{1}{2}+\frac{i\sqrt{3}}{2} \right)}^{365}}\]is equal to [IIT 1999]

    A) \[1-i\sqrt{3}\]

    B) \[-1+i\sqrt{3}\]

    C) \[i\sqrt{3}\]

    D) \[-i\sqrt{3}\]

    Correct Answer: C

    Solution :

    Given equation is \[4+5{{\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)}^{365}}\] \[=4+5{{\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)}^{334}}\]\[+3{{\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)}^{365}}\] \[=4+5\left[ \cos \frac{668}{3}\pi +i\sin \frac{668}{3}\pi  \right]\]\[3\left[ \cos \frac{730}{3}\pi +i\sin \frac{730}{3}\pi  \right]\] \[=4+5\left[ \cos \left( 222\pi +\frac{2\pi }{3} \right)+i\sin \left( 222\pi +\frac{2\pi }{3} \right) \right]\]\[+3\left[ \cos \left( 243\pi +\frac{\pi }{3} \right)+i\sin \left( 243\pi +\frac{\pi }{3} \right) \right]\] \[=4+5\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)+3\left( -\cos \frac{\pi }{3}-i\sin \frac{\pi }{3} \right)\] \[=4+5\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)+3\left( -\frac{1}{2}-i\frac{\sqrt{3}}{2} \right)\] \[=4-4+2i\frac{\sqrt{3}}{2}=i\sqrt{3}\].


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