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Complex Numbers and Quadratic Equations Test 35

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Complex Numbers and Quadratic Equations Test 35
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Weekly Quiz Competition
  • Question 1
    1 / -0
    z1z_1 and z2z_2 are two non-zero complex numbers such that z1=2+4iz2=56iz_1=2+4i\\z_2=5-6i, then z2z1z_2-z_1 equals
    Solution
    z1=2+4iz2=56iz2z1=56i24i=310iz_1=2+4i\\\\z_2=5-6i\\\\z_2-z_1=5-6i-2-4i=3-10i
  • Question 2
    1 / -0
    The imaginary part of t;tRt ; t \in R is 
    Solution

  • Question 3
    1 / -0
    If  z=cosπ4+isinπ6,z = \cos \dfrac { \pi } { 4 } + i \sin \dfrac { \pi } { 6 } ,  then
    Solution
    z=cosπ4+isinπ6z = \cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{6}

    z=12+i12z = \dfrac{1}{\sqrt{2}} + i \dfrac{1}{2}

    θ=tan11212=tan1(22)\theta = \tan^{-1} \dfrac{\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}} = \tan^{-1} \left(\dfrac{\sqrt{2}}{2} \right)

    θ=tan1(12)\theta = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right)

    argz=tan112\therefore arg z = \tan^{-1} \dfrac{1}{\sqrt{2}}
    z=(12)2+(12)2=12+14|z| = \sqrt{\left(\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{1}{2}\right)^2} = \sqrt{\dfrac{1}{2} + \dfrac{1}{4}}

    =2+14= \sqrt{\dfrac{2 + 1}{4}}

    =32= \dfrac{\sqrt{3}}{2}

    z=cosπ4+isinpi4z = \cos \dfrac{\pi}{4} + i \sin \dfrac{pi}{4}

    argz=angle=π4arg z = angle = \dfrac{\pi}{4}

    and z=cos2π4+sin2π4=1|z| = \cos^2 \dfrac{\pi}{4} + \sin^2 \dfrac{\pi}{4} = 1
  • Question 4
    1 / -0
    Let z be a complex number such that the principal value of argument, arg z<0z < 0. Then arg(z)arg(z)arg(-z) - arg (z) is
    Solution
    Consider
    z=zeiθz=|z|e^{iθ}

    Now,
    z=zei(θ±π)=zeiθˉ−z=|z|e^{i(θ±π)}=|z|e^{i\bar{\theta}}

    whereθˉ=arg(z) \bar{\theta}=arg(−z)

    Then
    arg(z)arg(z)=θˉθ=±πarg(−z)−arg(z)=\bar{\theta}−θ=±π

    This has been verified numerically for random values of zz.
  • Question 5
    1 / -0
    Given z1+3z24z3=0 z _ { 1 } + 3 z _ { 2 } - 4 z _ { 3 } = 0 then z1,z2,z3 z _ { 1 } , z _ { 2 } , z _ { 3 } are
    Solution

  • Question 6
    1 / -0
    If z be a complex number satisfying z4+z3+2z2+z+1=0z^{4}+z^{3}+2z^{2}+z+1=0 then z=\left|z\right|=
    Solution
    z4+z3+2z2+z+1=0z^{4}+z^{3}+2z^{2}+z+1=0
    z4+z2+z3+z+z2+1=0\Rightarrow z^{4}+z^{2}+z^{3}+z+z^{2}+1=0 by rearranging
    z2(z2+1)+z(z2+1)+1(z2+1)=0\Rightarrow z^{2}(z^{2}+1)+z(z^{2}+1)+1(z^{2}+1)=0
    z2(z2+1)+(z2+1)(z+1)=0\Rightarrow z^{2}(z^{2}+1)+(z^{2}+1)(z+1)=0
    (z2+1)(z2+z+1)=0\Rightarrow (z^{2}+1)(z^{2}+z+1)=0
    z2+1=0,z2+2+1=0\Rightarrow z^{2}+1=0, z^{2}+2+1=0
    z=±i, z=1±i32\Rightarrow z=\pm i,  z=\frac{-1\pm i\sqrt{3}}{2}
    z=i,i,1+i32,1i32\therefore z=i, -i, \frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}
    z=+1\therefore |z|=+1

  • Question 7
    1 / -0
    Express the following complex numbers in the standard form a+ib a+ib :
    (114i21+i)(34i5+i) \left ( \dfrac{1}{1-4i}-\dfrac{2}{1+i} \right )\left ( \dfrac{3-4i}{5+i} \right )
    Solution

  • Question 8
    1 / -0
    Express the following complex numbers in the standard form a+ib a+ib :
    (2+i)32+3i \dfrac{\left ( 2+i \right )^{3}}{2+3i}
    Solution

  • Question 9
    1 / -0
    Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
    1i1-i
    Solution
    Let z=1iz=1-i. Then, z=12+(1)2=2\left | z \right |=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}.

    Let α\alpha be the acute angle given by tanα=Im(z)Re(z)tan\, \alpha =\dfrac{\left | Im(z) \right |}{\left | Re(z) \right |}.

    Then,
    tanα=11=1α=π4tan\, \alpha =\dfrac{|-1|}{|1|}=1\Rightarrow \alpha =\dfrac{\pi }{4}

    Clearly, z lies in the fourth quadrant. Therefore, arg(z)=α=π4arg(z)= -\alpha =-\dfrac{\pi }{4}.
  • Question 10
    1 / -0
    Express the following complex numbers in the standard form a+ib a+ib :
    34i(42i)(1+i) \dfrac{3-4i}{\left ( 4-2i \right )\left ( 1+i \right )}
    Solution

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