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Complex Numbers and Quadratic Equations Test -7

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Complex Numbers and Quadratic Equations Test -7
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  • Question 1
    1 / -0
    If $$z=2-3i$$ then $$z^2-4z+13=$$
    Solution
    $$z=2-3i$$
    $$z^{2}=2^{2}-3^{2}-12i$$
    $$=-5-12i$$
    $$\therefore z^{2}-4z+13$$
    $$=(-5-12i)-4(2-3i)+13$$
    $$=-5-12i-8+12i +13$$
    $$=-13+13$$
    $$=0$$

  • Question 2
    1 / -0
    The complex number $$\displaystyle \frac{1+2i}{1-i}$$ lies in the quadrant :
    Solution
    Let  $$z =\dfrac{1+2i}{1-i}$$

    $$\Rightarrow z =\dfrac{(1+2i)}{1-i}\times \dfrac{1+i}{1+i}$$

    $$= \dfrac{1+2i+i+2i^2}{1-i^2}$$

    $$=\dfrac{1+3i-2}{1+1}$$ ............ $$[\because i^2 = -1]$$

    $$\Rightarrow z =\dfrac{-1+3i}{2}$$

    $$ =-\dfrac{1}{2}+\dfrac{3}{2}i $$
    $$=x+iy$$
    Clearly $$x<0$$ and $$y>0$$
    Hence $$z$$ lies in $$\text{II}$$ quadrant
  • Question 3
    1 / -0
    $$\sqrt{-3}\sqrt{-75}=$$
    Solution
    $$\sqrt{-3}\times\sqrt{-75}=\sqrt{3\times(-1)}\sqrt{75\times(-1)}$$
    $$=\sqrt{3}i\times\sqrt{75}i$$
    $$=\sqrt{225}i^{2}$$
    $$= - 15$$
  • Question 4
    1 / -0
    The sum of two complex numbers $$a + ib$$ and $$c +id$$ is a real number if
    Solution
    It is given that 

    $$z_{1}=a+ib$$ and 

    $$z_{2}=c+id$$

    Then
    $$z_{1}+z_{2}=(a+c)+i(b+d)$$

    Now
    $$(z_{1}+z_{2})$$ is purely real.

    Then the imaginary part has to be $$0$$.

    Hence
    $$b+d=0$$.
  • Question 5
    1 / -0
    The locus of complex number z such that z is purely real and real part is equal to - 2 is
    Solution
    $$z=x+iy$$
    $$(x,y)$$
    $$z$$ is purely real and the real part equals $$-2$$
    $$\therefore y=0$$ & $$x=-2$$
    $$z=-2$$
    Hence, this would be represented by the point (-2,0) on the Argand Plane.
  • Question 6
    1 / -0
    $$\dfrac{1}{i-1}+\dfrac{1}{i+1}$$ is
    Solution
    Let $$Z= \dfrac{1}{i-1}+\dfrac{1}{i+1}$$

             $$=\dfrac{i+1+i-1}{(i-1)(i+1)}$$

             $$=\dfrac{i+i}{(i^2-1^2)}$$

             $$=\dfrac{2i}{-2}$$

    $$ \therefore Z=-i$$
  • Question 7
    1 / -0
    If $$(x+iy)(2-3i)=4+i$$ then (x, y) =
    Solution
    $$(x+iy)(2-3i)=4+i$$

    $$2x-(3x)i+(2y)i-3yi^{2}=4+i$$

    $$\underbrace{2x+3y}_{Real }+\underbrace{(2y-3x)}_{Imaginary }i=4+i$$

    Comparing the real & imaginary parts,
    $$2x+3y=4$$--------------------------(1)
    $$2y-3x=1$$----------------------------(2)
    Solving eq(1)  &  eq(2),
    $$4x+6y=8$$
    $$-9x+6y=3$$
    $$13x=5\Rightarrow x=\dfrac{5}{13}$$
    $$y=\dfrac{14}{13}$$
    $$\therefore (x,y)=\left (\dfrac{5}{13},\dfrac{14}{13} \right )$$

  • Question 8
    1 / -0
    The argument of every complex number is
    Solution
    $$z=x+iy$$
    amplitude $$= tan^{-1}\frac{y}{x}$$
    $$\Rightarrow $$ amplitude $$=\theta \pm 2k\pi $$    where   $$\theta \epsilon \left [ -\pi ,\pi \right ]$$   $$\forall k\epsilon R$$
    since $$k\epsilon R$$
    $$\Rightarrow $$ Amplitude of any complex number is many valued.
  • Question 9
    1 / -0
    The sum of two complex numbers $$a + ib$$ and $$c+ id$$ is purely imaginary if
    Solution
    It is given that
    $$z_{1}=a+ib$$ and
    $$z_{2}=c+id$$
    $$z_{1}+z_{2}=(a+c)+i(b+d)$$
    $$z_{1}+z_{2}$$ is purely imaginary.       (Given)
    Then the real part has to be  $$0$$.
    Hence
    $$a+c=0$$.
  • Question 10
    1 / -0
    For $$a > 0$$, arg $$(ia) =$$
    Solution
    $$z= 0 + ia$$
    $$\therefore arg(z) = arg (ia) = \tan^{-1} \cfrac a0$$
      = $$\cfrac {\pi}{2}$$        ...(since a is greater than or equal to zero)
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