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Complex Numbers and Quadratic Equations Test 17

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Complex Numbers and Quadratic Equations Test 17
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  • Question 1
    1 / -0
    The product of $$(3-2i)$$ and $$\left(\dfrac { 5 }{ 2 } -4i\right)$$, if $$i=\sqrt { -1 } $$ , is:
    Solution
    $$(3 - 2i) \times \left(\dfrac{5}{2} - 4i\right)$$

    $$= 3\left(\dfrac{5}{2} - 4i\right) - 2i\left(\dfrac{5}{2} - 4i\right)$$

    $$= \dfrac{15}{2} - 12i - 5i - 8$$

    $$= \dfrac{-1}{2} - 17i$$
  • Question 2
    1 / -0
    The resultant complex number when $$(4+6i)$$ is divided by $$(10-5i)$$ is
    Solution
    The value of $$\cfrac{4 + 6i}{10 - 5i}$$ 
    $$= \cfrac{4 + 6i}{10 - 5i} \times \cfrac{10 + 5i}{10 + 5i}$$
    $$= \cfrac{(4 + 6i)(10 + 5i)}{(10 - 5i)(10 + 5i)}$$
    $$= \cfrac{40 + 20i + 60i - 30}{100 + 25}$$
    $$= \cfrac{10 + 80i}{125}$$
    $$= \cfrac{2}{25} + \cfrac{16}{25}i$$
  • Question 3
    1 / -0
    The argument of $$\dfrac {1 + i\sqrt {3}}{\sqrt {3} + i}$$ is
    Solution
    Let $$z = \dfrac {1 + i\sqrt {3}}{\sqrt {3} + i}$$
    $$= \dfrac {1 + i\sqrt {3}}{(\sqrt {3} + i)} \times \dfrac {(\sqrt {3} - i)}{(\sqrt {3} - i)}$$
    $$= \dfrac {\sqrt {3} - i + 3i + \sqrt {3}}{3 + 1} = \dfrac {\sqrt {3} + i}{2}$$
    $$\arg(z) = \tan^{-1}\left (\dfrac {1}{\sqrt {3}}\right )$$
    $$= \dfrac {\pi}{6}$$
  • Question 4
    1 / -0
    The principal amplitude of $$\displaystyle { \left( \sin { { 40 }^{ \circ  }+i\cos { { 40 }^{ \circ  } }  }  \right)  }^{ 5 }$$ is
    Solution
    $$\displaystyle { \left( \sin { { 40 }^{ \circ  } } +i\cos { { 40 }^{ \circ  } }  \right)  }^{ 5 }$$
    $$\displaystyle ={ i }^{ 5 }{ \left( \cos { { 40 }^{ \circ  } } -i\sin { { 40 }^{ \circ  } }  \right)  }^{ 5 }$$
    $$\displaystyle =i\left( \cos { { 200 }^{ \circ  } } -i\sin { { 200 }^{ \circ  } }  \right) $$
    $$\displaystyle =i\left[ \cos { \left( { 180 }^{ \circ  }+{ 20 }^{ \circ  } \right)  } -i\sin { \left( { 180 }^{ \circ  }+{ 20 }^{ \circ  } \right)  }  \right] $$
    $$\displaystyle =i\left( -\cos { { 20 }^{ \circ  } } +i\sin { { 20 }^{ \circ  } }  \right) $$
    $$\displaystyle =-i\cos { { 20 }^{ \circ  } } -\sin { { 20 }^{ \circ  } } $$
    $$\displaystyle =\cos { \left( -{ 11 }0^{ \circ  } \right) + } i\sin { \left( -{ 11 }0^{ \circ  } \right)  } $$
    $$\displaystyle \therefore $$ Principal amplitude = $$\displaystyle -{ 110 }^{ \circ  }$$.
  • Question 5
    1 / -0
    If $$A = (3 - 4i)$$ and $$B = (9 + ki)$$, where $$k$$ is a constant. 
    If $$AB - 15 = 60$$, then the value of $$k$$ is
    Solution

    Given, $$A=3-4i, B=9+ki, Ab-15=0$$

    $$\therefore AB= (3-4i)(9+ki)$$

    $$\therefore 27 + 3ki – 36i – 4ki^2-15=60$$

    $$\therefore -48+3ki-36i+4k=0$$

    Separate the real and imaginery part equal to zero, then we get the value of $$k$$,

    $$\therefore -48 + 4k = 0, 3k – 36 = 0$$

    $$\therefore -48 = -4k, 3k = 36$$

    $$\therefore k = 12, k = 12$$

    So, the value of $$ k$$ is $$12$$.

  • Question 6
    1 / -0
    The simplest form of $$\sqrt {-18} \times \sqrt {-50}$$ is
    Solution
    We know that $$i^2=-1$$
    So, $$\sqrt{18i^2}\times\sqrt{50i^2}$$
    $$=$$ $$i\sqrt{18}\times i\sqrt{50}$$
    $$=$$ $$i^2\sqrt{18\times50}$$
    $$=$$ $$-\sqrt{900}$$
    $$= - 30$$
  • Question 7
    1 / -0
    Simplify $$(2+8i)(1-4i)-(3-2i)(6+4i)$$
     (Note$$:i=\sqrt{-1}$$)
    Solution
    The value of $$\left ( 2+8i \right )\left ( 1-4i \right )-\left ( 3-2i \right )\left ( 6+4i \right )$$
    $$=$$ $$\left ( 2+8i-8i-32i^{2} \right )-\left ( 18-12i+12i-8i^{2} \right )$$
    $$=$$ $$2-32i^{2}-18+8i^{2}$$
    Put the given value of $$i=\sqrt{-1}$$, we get
    $$=$$ $$2-32(\sqrt{-1})^{2}-18+8(\sqrt{-1})^{2}$$
    $$=$$ $$2-32(-1)-18+8(-1)$$
    $$=$$ $$2+32-18-8$$
    $$=$$ $$34-26=8$$
  • Question 8
    1 / -0
    Given that $$4$$ is a root of the quadratic equation $${x}^{2}-5x+q=0$$. Find the value of $$q$$ and the other root.
  • Question 9
    1 / -0
    The imaginary number $$i$$ is defined such that $$i^2=-1$$. What is the value of $$(1 - i \sqrt {5}) ( 1 + i\sqrt {5})$$?
    Solution
    Imaginary number is $$i=\sqrt {-1}$$

    Value of $$(1-i\sqrt{5})(1+i\sqrt{5})$$

    It is in the form of $$(a+b)(a-b)=a^2-b^2$$

    So, $$(1-i\sqrt{5})(1+i\sqrt{5})={(1)}^2-{(i\sqrt {5})}^2$$

    $$\Rightarrow 1-(i^2\times 5)$$

    $$\Rightarrow 1-(-1\times 5)$$

    $$\Rightarrow 6$$

    $$(1-i\sqrt{5})(1+i\sqrt{5})=6$$
  • Question 10
    1 / -0
    If $$i = \sqrt {-1}$$, find the values of $$n$$ such that $$i^{n} + (i)^{n}$$ have a positive value.
    Solution
    $${ i }^{ n }+{ i }^{ n }=2{ i }^{ n }$$
    We know that $${ i }^{ 2 }=-1$$ and $${ i }^{ 4 }=1$$, so $$n$$ should be multiple of $$4$$.
    Among given options $$24$$ will be divisible by $$4$$.
    Therefore the answer is $$24$$.
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