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Number Theory Test 37

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Number Theory Test 37
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  • Question 1
    1 / -0
    Modulus of $$\dfrac{\cos \theta - i\sin \theta}{\sin \theta - i \cos \theta}$$ is
    Solution
    $$\cfrac { \cos { \theta  } -i\sin { \theta  }  }{ \sin { \theta  } -i\cos { \theta  }  } $$
    $$\cfrac { \cos { \theta  } -i\sin { \theta  }  }{ \sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  }  } \left( \sin { \theta  } +i\cos { \theta  }  \right) \Rightarrow \left( \cos { \theta  } -i\sin { \theta  }  \right) \left( \sin { \theta  } +i\cos { \theta  }  \right) $$
    $$\Rightarrow \cos { \theta  } .\sin { \theta  } +\sin { \theta  } .\cos { \theta  } +i\cos ^{ 2 }{ \theta  } -i\sin ^{ 2 }{ \theta  } \Rightarrow 2\sin { \theta  } .\cos { \theta  } +i\left( \cos ^{ 2 }{ \theta  } -\sin ^{ 2 }{ \theta  }  \right) $$
    $$2\sin { \theta  } .\cos { \theta  } =\sin { 2\theta  } ;\cos ^{ 2 }{ \theta  } -\sin ^{ 2 }{ \theta  } =\cos { 2\theta  } \Rightarrow \sin { 2\theta  } +i\cos { 2\theta  } $$
    modulus $$=\sqrt { \sin ^{ 2 }{ 2\theta  } +\cos ^{ 2 }{ 2\theta  }  } =\sqrt { 1 } =1$$
    none of these
  • Question 2
    1 / -0

    The value of $$\sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)} $$, where $$i = \sqrt { - 1} $$ equals:

    Solution

  • Question 3
    1 / -0
    $$n\in N,\ { \left( \dfrac { 1+i }{ \sqrt { 2 }  }  \right)  }^{ 8n }+{ \left( \dfrac { 1-i }{ \sqrt { 2 }  }  \right)  }^{ 8n }=$$
    Solution

    We have,

    $${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{8n}}+{{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{8n}}$$

    $$ {{\left[ {{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{2}} \right]}^{4n}}+{{\left[ {{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{2}} \right]}^{4n}} $$

    $$ ={{\left[ \dfrac{{{1}^{2}}+{{i}^{2}}+2i}{2} \right]}^{4n}}+{{\left[ \dfrac{{{1}^{2}}+{{i}^{2}}-2i}{2} \right]}^{4n}} $$

    $$ ={{\left[ \dfrac{1-1+2i}{2} \right]}^{4n}}+{{\left[ \dfrac{1-1-2i}{2} \right]}^{4n}} $$

    $$ ={{\left( {{i}^{4}} \right)}^{n}}+{{\left( {{\left( -i \right)}^{4}} \right)}^{n}} $$

    $$ ={{1}^{n}}+{{1}^{n}} $$

    $$ =1+1 $$

    $$ =2 $$


    Hence, this is the answer.
  • Question 4
    1 / -0
    $$\left(\dfrac{1+\cos \dfrac{\pi}{8}+i\sin \dfrac{\pi}{8}}{1+\cos \dfrac{\pi}{8}-i\sin \dfrac{\pi}{8}}\right)^{8}=$$ ?
    Solution

  • Question 5
    1 / -0
    $$3+2\ i\ \sin \theta$$ will be real, if $$\theta=$$
    Solution
    $$3+2i\sin \theta$$ will be real if imaginary term is zero
    i.e $$2\sin\theta=0$$
    $$\Rightarrow \sin \theta=0$$
    $$\Rightarrow \theta=n\pi$$
  • Question 6
    1 / -0
    Let $$z_{r}(1 \le r \le 4)$$ be complex numbers such that $$|z_{r}|=\sqrt {r+1} and |30\ z_{1}+20\ z_{2}+15 z_{3}+12\ z_{4}|=k|z_{1}z_{2}z_{3}+z_{2}z_{3}z_{4}+z_{3}z_{4}z_{1}+z_{4}z_{1}z_{2}|$$. Then value of $$k$$ equals ?
    Solution
    $$|z_1| = \sqrt 2, \ |z_2| = \sqrt 3, \ |z_3| = \sqrt 4, \ |z_4| = \sqrt 5$$

    $$ | 30z_1 + 20 z_2 + 15 z_3 + 12 z _4 | = k | z_1z_2z_3 + z_2z_3z_4+z_3z_4z_1+z_4z_1z_2| $$

    $$ \overline{|30z_1 + 20 z _2 + 15 z_3 + 12 z_4|} =k | z_1z_2z_3+z_2z_3z_4+z_3z_4z_1+z_4z_1z_2|$$

    $$ | 30 \bar {z_1} + 20 \bar {z_2} + 15 \bar {z_3} + 12 \bar {z_4} | = k|z_1z_2z_3+z_2z_3z_4+z_3z_4z_1+z_4z_1z_2$$

    $$ z_1 \bar{z_1} = 2, \ z_2 \bar { z_2} = 3 , \ z_3 \bar { z_3} = 4, \ z_4\bar{z_4} = 5$$

    $$60 \left | \dfrac{ 1}{z_1} + \dfrac{1}{z_2} + \dfrac{1}{z_3} + \dfrac{1}{z_4} \right | = k |z_1z_2z_3+z_2z_3z_4+z_3z_4z_1+z_4z_1z_2|$$

    $$k = |z_4z_1z_2|$$
  • Question 7
    1 / -0
    The argument of the complex number $$\sin \dfrac {6\pi}{5}+i\left(1+\cos \dfrac {6\pi}{5}\right)$$ is
    Solution
    $$\sin{\dfrac{6\pi}{5}}+i\left(1+\cos{\dfrac{6\pi}{5}}\right)$$ is equal to $$\sin{\left(\dfrac{\pi}{5}\right)}+i\left(1-\cos{\left(\dfrac{\pi}{5}\right)}\right)$$
    Lies in the second quardent of complex plane and its argument is
    $$arg \left(n+iy\right)=\pi-{\tan}^{-1}{\left|\dfrac{y}{x}\right|}$$
    $$=\pi-{\tan}^{-1}{\left|\dfrac{1-\cos{\left(\dfrac{\pi}{5}\right)}}{ \sin{\left(\dfrac{\pi}{5}\right)}}\right|}$$
    $$=\pi-{\tan}^{-1}{\left|\dfrac{2{\sin}^{2}{\left(\dfrac{\pi}{10}\right)}}{ 2\sin{\left(\dfrac{\pi}{10}\right)\cos{\left(\dfrac{\pi}{10}\right)}}}\right|}$$
    $$=\pi-{\tan}^{1}{\left|\dfrac{\sin{\left(\dfrac{\pi}{10}\right)}}{\cos{\left(\dfrac{\pi}{10}\right)}}\right|}$$
    $$=\pi-{\tan}^{-1}{\left|\tan{\left(\dfrac{\pi}{10}\right)}\right|}$$
    $$=\pi-{\tan}^{-1}{\left(\tan{\left(\dfrac{\pi}{10}\right)}\right)}$$
    $$=\pi-\dfrac{\pi}{10}$$
    $$=\dfrac{9\pi}{10}$$

  • Question 8
    1 / -0

    For a complex number $$z$$, the minimum value of $$\left| z \right| + \left| {z - 1} \right|$$ is

    Solution

  • Question 9
    1 / -0
    If $$\left| {{z_1}} \right| =  = 1,\left| {{z_2}} \right| = 2,$$, then the value of $${\left| {{z_1} + {z_2}} \right|^2} + {\left| {{z_1} - {z_2}} \right|^2}$$ is equal to 
    Solution
    $$ | z _1 + z _2 | ^2 + |z _1-z_2|^2 =  (z _1+z_2 ) ( \bar{z_1} + \bar {z_2}) + (z_1-z_2)(\bar{z_1} - \bar {z_2})$$

    $$\Rightarrow 2z_1 \bar{z_1} + 2 z_2 \bar {z_2} =  2 ( |z_1|^2 + |z_2|^2) = 2 ( 1 + 4 ) = 10$$ 
  • Question 10
    1 / -0
    The modulus of the complex quantity $$(2-3i)(-1+7i)$$.
    Solution
    The given complex quantity is $$(2-3i)(-1+7i)$$.

    Let $$z_{1}=2-3i$$ and $$z_{2}=-1+7i$$

    Therefore, $$|z_{1}|=\sqrt{2^2+(-3)^2}=\sqrt{13}$$

    $$|z_{2}|=\sqrt{(-1)^2+7^2}=\sqrt{50}=5\sqrt{2}$$

    Thus, required modulus = $$|z_{1}z_{2}|=\sqrt{13} \times 5\sqrt{2}=5\sqrt{26}$$.
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