Mathematics

Complex Variables and Numbers

157 Questions

Complex numbers and variables form a crucial part of advanced mathematics syllabi. This topic covers roots of unity, Argand plane geometry, and Z-transforms. These concepts are frequently tested in engineering entrance exams and UPSC mathematics optional papers.

Roots of unityArgand plane geometryZ-transform sequencesExponential complex formsComplex number quadrantsRiemann hypothesis

Complex Variables and Numbers Questions

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

The radius of the locus by the point represented by $z$, when $arg\dfrac {z-1}{z+1} =\dfrac {\pi}{4}$, is

  1. $\sqrt {2}$
  2. $\sqrt {2}\pi$
  3. $\dfrac {\pi}{\sqrt {2}}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Arg$\left[\cfrac{z-1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$Arg$\left[{z-1}\times \cfrac{1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$Arg$\left[{z-1}\right] \times$Arg$\left[\cfrac{1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$ Arg${\left[z-1\right]}-$Arg${\left[z+1\right]}=\dfrac{\pi}{4}$

Let $z=x+iy$ 

$\therefore$Arg$\left[\left(x-1\right)+iy\right]-$Arg$\left[\left(x+1\right)+iy\right]=\dfrac{\pi}{4}$

$\Rightarrow {\tan}^{-1}\left(\dfrac{y}{x-1}\right)-{\tan}^{-1}\left(\dfrac{y}{x+1}\right)=\dfrac{\pi}{4}$

$\Rightarrow {\tan}^{-1}{\left(\dfrac{\dfrac{y}{x-1}-\dfrac{y}{x+1}}{1+\dfrac{{y}^{2}}{{x}^{2}-1}}\right)}=\dfrac{\pi}{4}$

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

$\Rightarrow {x}^{2}-1+{y}^{2}=2y$

$\Rightarrow {x}^{2}-2y+{y}^{2}-1=0$

This represents a circle with center at $\left(0,1\right)$ and radius  $=\sqrt{0+1-\left(-1\right)}=\sqrt{2}$
Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

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 

  1. $|z-7-4i|=1$
  2. $|z-7-5i|=\sqrt {2}$
  3. $|z-4i|=8$
  4. $|z-7i|=\sqrt {18}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation arg((z-z1)/(z-z2)) = pi/4 represents an arc of a circle passing through z1 and z2. Given z1=8+4i and z2=6+4i, the midpoint is 7+4i. The geometry of the angle pi/4 implies a specific circle equation.

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

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 

  1. 2i

  2. -2

  3. -2i

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. $2\sqrt{3}+2i$
  2. $2\sqrt{3}-2i$
  3. $-2\sqrt{3}+2i$
  4. $-\sqrt{3}+i$
Reveal answer Fill a bubble to check yourself
C Correct answer
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, $\Rightarrow z=-2\sqrt 3+2i$

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

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

  1. (-2,-1)

  2. (-2,1)

  3. (2,-1)

  4. (2,1)
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(-2,-1)$

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

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

  1. $\dfrac{{6\pi }}{5}$
  2. $\dfrac{{5\pi }}{5}$
  3. $\dfrac{{9\pi }}{10}$
  4. $\dfrac{{7\pi }}{10}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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) $

$arg(z)=\cfrac{9\pi}{10}$
Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

Let $A$ and $B$ represent $z _{1}$ and $z _{2}$ in the Argand plane and $z _{1},z _{2}$ be the roots of the equation $z^{2}+pz+q=0$ where $p,q$ are complex numbers. If $O$ is the origin $OA=OB$ and $\angle AOB=\alpha$ then $p^{2}=$

  1. $2q\ \cos \left(\dfrac{\alpha}{2}\right)$
  2. $4q\ \cos \left(\dfrac{\alpha}{2}\right)$
  3. $4q\ \cos^{2} \left(\dfrac{\alpha}{2}\right)$
  4. $4q^{2}\ \cos^{2} \left(\dfrac{\alpha}{2}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

If Arg $(z + i)\, -$ Arg $(z - i)$ $= \dfrac{\pi}{2}$, then $z$ lies on a ..........

  1. Circle

  2. Line

  3. Coordinate axes

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Putting z = x+iy,

${tan}^{-1}\dfrac{y+1}{x}$  -  ${tan}^{-1}\dfrac{y-1}{x}$ = $\pi$/2

$\Rightarrow$ 1 + ($\dfrac{y+1}{x})$($\dfrac{y-1}{x}$) = 0

$\Rightarrow$ $x^2 +  y^2$ = 1

It is a circle of center coinciding with origin and radius 1 units.

Hence, option A is correct.