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Complex Numbers and Quadratic Equations Test 10

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Complex Numbers and Quadratic Equations Test 10
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  • Question 1
    1 / -0
    Discuss the nature of the roots of the equations  (a+cb)x2+2cx+(b+ca)=0\displaystyle (a+c-b)x^{2}+2cx+(b+c-a)=0
    Solution
    (a+cb)x2+2cx+(b+ca)=0D=B24AC=(2c) 24(b+ca)(a+cb)=4c24(ab+cbb2+ac+c2cba2ac+ab)=4c2+4b2+4a24c28ab=4a2+4b28ab=4(ab) 2>0\left( a+c-b \right) { x }^{ 2 }+2cx+\left( b+c-a \right) =0\\ D={ B }^{ 2 }-4AC\\ ={ \left( 2c \right)  }^{ 2 }-4\left( b+c-a \right) \left( a+c-b \right) \\ =4{ c }^{ 2 }-4\left( ab+cb-{ b }^{ 2 }+ac+{ c }^{ 2 }-cb-{ a }^{ 2 }-ac+ab \right) \\ =4c^{ 2 }+4{ b }^{ 2 }+{ 4a }^{ 2 }-{ 4c }^{ 2 }-8ab\\ ={ 4a }^{ 2 }+{ 4b }^{ 2 }-8ab=4{ \left( a-b \right)  }^{ 2 }>0
    Hence roots  are real and unequal
  • Question 2
    1 / -0
    If 2-2 is a root of the quadratic equation x2+px+2=0x^2 + px + 2 = 0 and the quadratic equation 2x2+px+k=02x^2 + px+ k = 0 has equal roots, find the value of kk.
    Solution
    Given, 2 -2 is one of the root of first equation x2+px+2=0 x^2 + px + 2 = 0
    Let's take α \alpha as other root of the equation. 
    Now, product of the roots is given by (2).α=2 (-2).\alpha = 2 \longrightarrow α=1 \alpha = -1 and from sum of roots we get 2+(1)=p -2 + (-1) = -p \longrightarrow p=3 p = 3
    Now in the second equation roots are equal, take it as β\beta.
    From sum of roots we get 2.β=p2 2.\beta = \dfrac{-p}{2} \longrightarrow β=34\beta = \dfrac{-3}{4} and from product of the roots we get β2=k2{\beta}^{2} = \dfrac{k}{2} 
    \Rightarrow k=2.β2 k = 2.{\beta}^{2}  
    \Rightarrow k=98 k = \frac{9}{8} 
  • Question 3
    1 / -0
    If z=reiθz = re^{i\theta}, then the value of eiz|e^{iz}| is equal to
    Solution
    Ginen,
    z=reiθ  z=r{ e }^{ i\theta  }
    To findeiz \left| { e }^{ iz } \right|
     solution, 
    for given,z=reiθ z=r(cosθ+isinθ){eiθ =(cosθ+isinθ)}iz=ir(cosθ+isinθ)iz=r(icosθ+i2sinθ)iz=r(sinθ+icosθ)iz=rsinθ+ircosθeiz=ersinθ+ircosθ eiz=ersinθ eircosθ z=r{ e }^{ i\theta  }\\ z=r(\cos\theta +i\sin\theta )\left\{ { e }^{ i\theta  }=(\cos\theta +i\sin\theta ) \right\} \\ iz=ir(\cos\theta +i\sin\theta )\\ iz=r(i\cos\theta +{ i }^{ 2 }\sin\theta )\\ iz=r(-\sin\theta +i\cos\theta )\\ iz=-r\sin\theta +ir\cos\theta \\ { e }^{ iz }={ e }^{ -r\sin\theta +ir\cos\theta  }\\ \left| { e }^{ iz } \right| =\left| { e }^{ -r\sin\theta  } \right| \left| { e }^{ ir\cos\theta  } \right|
     We know thateiz=1eiz=ersinθ {alwayspositive}eiz=ersinθ  \left| { e }^{ iz } \right| =1\\ \therefore \left| { e }^{ iz } \right| =\left| { e }^{ -r\sin\theta  } \right| \longrightarrow \{ always\quad positive\} \\ { e }^{ iz }={ e }^{ -r\sin\theta  }
  • Question 4
    1 / -0
    Find the value of pp for which the quadratic equation x2+p(4x +p1)+2 =0x^2 + p(4x + p - 1) + 2 = 0 has equal roots ?
    Solution
    Given equation is x2+p(4x+p1)+2=0 x^2 + p(4x + p - 1) + 2 = 0.
    Comparing it with the ax2+bx+cax^2+bx+c, we get 
    a=1,b=4p,c=(p2p+2)a=1, b=4p, c=(p^2-p+2)
    Therefore, D=b24acD=b^2-4ac   
    x2+p(4x+p1)+2=0 x^2 + p(4x + p - 1) + 2 = 0.
    x2+4xp+p2p+2=0\Rightarrow x^2+4xp+p^2-p+2=0
    (4p)24(p2p+2)=0\Rightarrow (4p)^2-4(p^2-p+2)=0
    16p24p2+4p8=0\Rightarrow 16p^2-4p^2+4p-8=0
    12p2+4p8=0\Rightarrow 12p^2+4p-8=0
    3p2+p2=0\Rightarrow 3p^2+p-2=0
    3p2+32p2=0\Rightarrow 3p^2+3-2p-2=0
    3p(p+1)2(p+1)=0\Rightarrow 3p(p+1)-2(p+1)=0
    (p+1)(3p2)=0\Rightarrow (p+1)(3p-2)=0
    p=1\Rightarrow p=-1 and  p=23p=\dfrac {2}{3}
  • Question 5
    1 / -0
    Find the modulus and amplitude of 2i-2i
    Solution
    z=2iz=-2i            z=2|z|=2

    =2(i)=2(-i)

    =2eiπ2=2e^{i\frac{-\pi}{2}}

    =z(eiarg(z))=|z|(e^{i\arg(z)})

    Hence

    z=2|z|=2 and arg(z)=π2arg(z)=\dfrac{-\pi}{2}
  • Question 6
    1 / -0
    Find the modulus and amplitude of 2+23i-2 + 2 \sqrt 3i
    Solution
    Given z=2+23iz=-2+2\sqrt{3}i
    Hence, z=21+3 24|z|=2\sqrt{1+3}\ \Rightarrow 2\sqrt{4}
    2×2=4\Rightarrow 2\times2=4
    And let α\alpha be the argument of the complex number.
    Then
    tanα=3tan\alpha=-\sqrt{3}
    Now Re(z)<0Re(z)<0 and Im(z)>0Im(z)>0
    \therefore   the number lies in the II quadrant 
    Hence,
    α=ππ3\alpha=\pi-\dfrac{\pi}{3}

    =2π3=\dfrac{2\pi}{3}
  • Question 7
    1 / -0
    Find the modulus and amplitude of 3i-\sqrt 3-i
    Solution
    z=3iz=-\sqrt{3}-i    z=2|z|=2

    =2(3i2)=2(\dfrac{-\sqrt{3}-i}{2})

    =2ei(π6π)=2e^{i(\frac{\pi}{6}-\pi)}

    =2ei5π6=2e^{i\frac{-5\pi}{6}}

    =z.eiarg(z)=|z|.e^{iarg(z)}

    Hence

    z=2|z|=2 and arg(z)=5π6arg(z)=\dfrac{-5\pi}{6}.
  • Question 8
    1 / -0

    The amplitude of

    sinπ5+i(1cosπ5)\displaystyle sin \frac{\pi}{5} + i \left ( 1 - cos \frac{\pi}{5} \right )

    Solution
    Given,
    sinπ 5+i(1cosπ 5 )\Rightarrow\displaystyle \sin\frac { \pi  }{ 5 } +i\left( 1-\cos\frac { \pi  }{ 5 }  \right)

    =2sinπ 10cosπ 10+i2sin2π 10 =2\sin\dfrac { \pi  }{ 10 } \cos\dfrac { \pi  }{ 10 } +i2\sin^{ 2 }\dfrac { \pi  }{ 10 }   [sin2θ=2sinθcosθ2sin2θ=1cosθ][\because \sin2\theta=2\sin\theta \cos\theta\quad 2\sin^2\theta=1-\cos\theta]

     =2sinπ 10(cosπ 10+isinπ 10 )\displaystyle =2\sin\frac { \pi  }{ 10 } \left( \cos\frac { \pi  }{ 10 } +i\sin\frac { \pi  }{ 10 }  \right)

    Then

    Amplitude=tanθ=yxAmplitude=\tan\theta=\dfrac yx

    tanθ=sinπ 10 cosπ 10 =tanπ 10\displaystyle \tan\theta =\frac { \sin\frac { \pi  }{ 10 }  }{ \cos\frac { \pi  }{ 10 }  } =\tan\frac { \pi  }{ 10 }

    θ=π 10 \Rightarrow \theta =\dfrac { \pi  }{ 10 }
  • Question 9
    1 / -0
    If z1,z2,εCz_1, z_2, \varepsilon C are such that z1+z22=z12+z22|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 then z1z2\displaystyle \frac{z_1}{z_2} is
    Solution
    z1+z22=z12+z22\because |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2

    z12+z22+2R(z1zˉ2)=z12+z22\Rightarrow |z_1|^2 + |z_2|^2 + 2 R (z_1 \bar z_2) = |z_1|^2 + |z_2|^2

    R(z1zˉ2)=0z1zˉ2\Rightarrow R(z_1 \bar z_2) = 0 \Rightarrow z_1 \bar z_2 is imaginary

    Now z1z2=z1zˉ2z2zˉ2=imaginaryz22=\displaystyle \frac{z_1}{z_2} = \frac{z_1 \bar z_2}{z_2 \bar z_2} = \frac{\text{imaginary}}{|z_2|^2} = imaginary
  • Question 10
    1 / -0
    Find the value of x3+7x2x+16x^3 + 7x^2 -x + 16, where x=1+2ix = 1 + 2i
    Solution
    f(x)=x3+7x2x+16f\left(x\right)={x}^{3}+7{x}^{2}-x+16
    f(1+2i)=(1+2i)3+7(1+2i)2(1+2i)+16f(1+2i)=(1+2i)^3+7(1+2i)^2-(1+2i)+16
    =[1+8i3+6i(1+2i)]+7[1+4i2+4i]12i+16=\left[1+8{i}^{3}+6i\left(1+2i\right)\right]+7\left[1+4{i}^{2}+4i\right]-1-2i+16
    =[18i+6i12]+7[14+4i]2i+15=\left[1-8i+6i-12\right]+7\left[1-4+4i\right]-2i+15
    =[112i]+7[4i3]2i+15=\left[-11-2i\right]+7\left[4i-3\right]-2i+15
    =112i+28i212i+15=-11-2i+28i-21-2i+15
    =(32+15)+24i=\left(-32+15\right)+24i
    =17+24i=-17+24i

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