Tag: complex numbers and linear inequations

Questions Related to complex numbers and linear inequations

If $z _{1}=8 +4i,\ z _{2}=6+4i$ and $arg \left(\dfrac {z-z _{1}}{z-z _{2}}\right)=\dfrac {\pi}{4}$, then $z$ satisfy 

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $|z-7-4i|=1$

  2. $|z-7-5i|=\sqrt {2}$

  3. $|z-4i|=8$

  4. $|z-7i|=\sqrt {18}$

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Correct Option: A

In the complex plane, what is the distance of $4-2i$ from the origin?

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $2$

  2. $3.46$

  3. $4.47$

  4. $6$

  5. $12$

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Correct Option: C
Explanation:

Let $z=4-2i$

In a complex plane, distance of $z$ from any point origin is given by $|z|=\sqrt{(x _1)^2+(y _1)^2}$
$\therefore$ Distance of $z=4-2i$ from origin is given by $|z|=|4-2i|=\sqrt{4^2+(-2)^2}=\sqrt{20}\approx 4.47$
Hence, the answer is $4.47$.

In the complex plane, the number 4 + j3 is located in the

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. first quadrant

  2. second quadrant

  3. third quadrant

  4. fourth quadrant

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Correct Option: A
Explanation:

Since both the real and imaginary parts are positive , 4 + j3 lies in first quadrant of argand/complex plane.

If ${z _1}$ and ${z _2}$ are two non-zero complex number such that $\left| {{{{z _1}} \over {{z _2}}}} \right|$ = 2 and $\arg \left( {{z _1}{z _2}} \right) = {{3\pi } \over 2}$ , then ${{\overline {{z _1}} } \over {{z _2}}}$ is equal to 

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. 2i

  2. -2

  3. -2i

  4. 2

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Correct Option: A
Explanation:

$Let\quad { z } _{ 1 }=2r{ e }^{ i{ \theta  } _{ 1 } },{ z } _{ 2 }=r{ e }^{ i{ \theta  } _{ 2 } }\ arg\left( { z } _{ 1 }{ z } _{ 2 } \right) ={ \theta  } _{ 1 }+{ \theta  } _{ 2 }=\cfrac { 3\pi  }{ 2 } \ \bar { { z } _{ 1 } } =2r{ e }^{ -i{ \theta  } _{ 1 } }\ \cfrac { \bar { { z } _{ 1 } }  }{ { z } _{ 2 } } =\cfrac { 2r{ e }^{ -i{ \theta  } _{ 1 } } }{ r{ e }^{ i{ \theta  } _{ 2 } } } =2r{ e }^{ -i({ \theta  } _{ 1 }+{ \theta  } _{ 2 }) }=2r{ e }^{ -i\cfrac { 3\pi  }{ 2 }  }=2i$

Given $\left| z \right| =4$ and $Argz=\dfrac{5z}{6}$, then $z$ is

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $2\sqrt{3}+2i$

  2. $2\sqrt{3}-2i$

  3. $-2\sqrt{3}+2i$

  4. $-\sqrt{3}+i$

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Correct Option: C
Explanation:

Given that


$\Rightarrow |z|=4$ and $Arg \space z=\dfrac{5\pi}{6}$

$\Rightarrow $Let $z=x+iy$

$\Rightarrow |z|=4$ 

$\Rightarrow \sqrt{x^2+y^2}=4$

$\Rightarrow {x^2+y^2}=16$                ...$(1)$

$\Rightarrow Arg \space z=\dfrac{5\pi}{6}$

$\Rightarrow \tan ^{-1}(\dfrac{y}{x})=\dfrac{5\pi}{6}$

$\Rightarrow \dfrac{y}{x}=\tan (\dfrac{5\pi}{6})$

$\Rightarrow \dfrac{y}{x}=-\dfrac{1}{\sqrt 3}$
  
$\Rightarrow x^2=3y^2$

Substituting this in $(1)$ we get,

$\Rightarrow x^2+\dfrac{x^2}{3}=16$

$\Rightarrow \dfrac{4x^2}{3}=16$

$\Rightarrow x^2=12$

$\Rightarrow x=\pm 2\sqrt 3$

$\Rightarrow y=\mp 2$

$\Rightarrow \theta$ lies in $2^{nd} Quadrant$

Hence, $\Rightarrow z=-2\sqrt 3+2i$

$|z-4| < |z-2|$ represents the region given by?

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $Re(z) > 3$

  2. $Re(z) < 0$

  3. $Re(z) > 2$

  4. None of these

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Correct Option: A

If $a, b \notin R$, then $|e^{a + ib}| $ is equal to


maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $e^a$

  2. $e^b$

  3. $1$

  4. None of these

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Correct Option: A
Explanation:

$|e^{a + ib}| = |e^a . e^{ib}|$


            $= |e^a . (\cos b + i \sin b)|$


            $= e^a |(\cos b + i \sin b)|$

            $= e^a . \sqrt{(\cos b)^2 + (\sin b)^2}$

            $= e^a . \sqrt{1}$

            $= e^a$

If $Re(\dfrac{z+2i}{z+4})=0$ then z lies on a circle with center:

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. (-2,-1)

  2. (-2,1)

  3. (2,-1)

  4. (2,1)

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Correct Option: A
Explanation:

Let $z=x+ iy$

Then 

$w=\dfrac{z+2i}{z+4}=\dfrac{x+i(2+y)}{(x+4)+iy}$
Rationalizing we get:

$w=\dfrac{z+2i}{z+4}=\dfrac{x+i(2+y)}{(x+4)-iy}\times \dfrac{x+4-iy}{x+4+iy}=\dfrac{x^2+4x+y^2+2y+i(xy+2x+4y+xy+8)}{(x+4)^2+y^2}$
Since real part of $w$ is 0, hence

$(x+2)^2+(y+1)^2-5=0$

Hence centre of circle is $(-2,-1)$

The argument of the complex number $\sin \dfrac{{6\pi }}{5} + i\left( {1 + \cos \dfrac{{6\pi }}{5}} \right)$ is 

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $\dfrac{{6\pi }}{5}$

  2. $\dfrac{{5\pi }}{5}$

  3. $\dfrac{{9\pi }}{10}$

  4. $\dfrac{{7\pi }}{10}$

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Correct Option: C
Explanation:
Given that

$\sin { \cfrac { 6\pi  }{ 5 }  } +i\left( 1+\cos { \cfrac { 6\pi  }{ 5 }  }  \right) $

Notice the point $\sin { \left( \cfrac { 6\pi  }{ 5 }  \right)  } +i\left( 1+\cos { \cfrac { 6\pi  }{ 5 }  }  \right) $ or

$-\sin { \left( \cfrac { 6\pi  }{ 5 }  \right)  } +i\left( 1-\cos { \cfrac { \pi  }{ 5 }  }  \right) $ lies in the second quadrant of complex plane hence its argument is given as

$arg\left( z \right) =\pi -\tan ^{ -1 }{ \left| y/x \right|  } \quad $  ($\because z=x+iy$)

$\left( \forall x<0,y\ge 0 \right) $

$arg(z)=\pi -\tan ^{ -1 }{ \left| \cfrac { 1-\cos { \cfrac { \pi  }{ 5 }  }  }{ \sin { \cfrac { \pi  }{ 5 }  }  }  \right|  } $

$=\pi -\tan ^{ -1 }{ \left| \cfrac { 2\sin ^{ 2 }{ \cfrac { \pi  }{ 10 }  }  }{ 2\sin { \cfrac { \pi  }{ 10 }  } \cos { \cfrac { \pi  }{ 10 }  }  }  \right|  } =\pi -\tan ^{ -1 }{ \left| \cfrac { \sin { \cfrac { \pi  }{ 10 }  }  }{ \cos { \cfrac { \pi  }{ 10 }  }  }  \right|  } $

$=\pi -\tan ^{ -1 }{ \left| \tan { \cfrac { \pi  }{ 10 }  }  \right|  } \left( \because \tan ^{ -1 }{ \cfrac { \pi  }{ 10 }  } >1 \right) $

$=\pi -\cfrac { \pi  }{ 10 } \left( \because -\cfrac { \pi  }{ 2 } \le \tan ^{ -1 }{ x } \le \cfrac { \pi  }{ 2 }  \right) $

$arg(z)=\cfrac{9\pi}{10}$

Let $z,w$ be complex numbers such that $\vec {z}+i\vec {w}=$ and $zw=\pi$ Then $arg\ z$ equals

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $\dfrac {\pi}{4}$

  2. $\dfrac {5\pi}{4}$

  3. $\dfrac {3\pi}{4}$

  4. $\dfrac {\pi}{2}$

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Correct Option: C