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Complex Numbers and Quadratic Equations Test 30

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Complex Numbers and Quadratic Equations Test 30
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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
    If $$\cfrac{\pi}{3}$$ and $$\cfrac{\pi}{4}$$ are the arguments of $${z}_{1}$$ and $${\overline { z }  }_{ 2 }$$, then the value of arg $$({z}_{1}{z}_{2})$$ is
    Solution
    $$arg(z_1.z_2)=arg.(z_1)+arg(z_2)$$
    $$=\dfrac{\pi}{3}+\left(\dfrac{-\pi}{4}\right)$$
    $$\dfrac{\pi}{12}$$



















  • 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
    The complex number $$z$$ satisfies $$z+|z|=2+8i$$. The value of $$|z|$$ is
    Solution
    $$Assume,z=a+ib,where \ i=\sqrt { -1 } $$

    $$Then,z+|z|=a+ib+\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =2+8i$$

    $$Thus,b=8.$$

    $$And\quad a+\sqrt { { a }^{ 2 }+64 } =2$$

    $${ a }^{ 2 }+64={ (2−a) }^{ 2 }$$

    $${ a }^{ 2 }+64=4−4a+{ a }^{ 2 }$$

    $$a=−15$$

    $$Thus\quad z=−15+8i$$

    $$Thus|z|=\sqrt { { (-15) }^{ 2 }+{ 8 }^{ 2 } } =17$$
  • Question 5
    1 / -0

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

    Solution

  • Question 6
    1 / -0

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

    Solution

  • Question 7
    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 8
    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 = $$|z_{1}z_{2}|=\sqrt{13} \times 5\sqrt{2}=5\sqrt{26}$$.
  • Question 9
    1 / -0
    If A and B be two complex numbers satisfying $$\dfrac{A}{B}+\dfrac{B}{A}=1$$. Then the two points represented by A and B and the origin form the vertices of
    Solution
    Given A and B are 2 complex numbers and $$ \dfrac{A}{B}+\dfrac{B}{A} -1 $$
    Let $$ A = z_1$$
    $$ B =z_2$$
    $$ z_3 = 0 $$
    $$ \dfrac{z_1}{z_2}+\dfrac{z_{2}}{z_1} = 1 \Rightarrow z_{1}^{2}+z_{2}^{2} =z_{1}z_{2}$$
    $$ \Rightarrow z_{1}^{2}+z_{2}^{2} + z_{3}^{2} = z_1z_2 + z_2 z_3 + z_3 z_1 $$
    This means $$ z_1, z_2, z_3 $$ from an equilateral triangle

  • Question 10
    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 $$2\sin\theta=0$$
    $$\Rightarrow \sin \theta=0$$
    $$\Rightarrow \theta=n\pi$$
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