Tag: complex numbers and linear inequations

Quizzes (4) Questions (127)

Questions Related to complex numbers and linear inequations

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

Let  $z _ { 1 } , z _ { 2 }$  and  $z _ { 3 }$  represent the vertices  $A, B$  and  $C$  of the triangle  $A B C$  in the argand that  $\left| z _ { 1 } \right| = \left| z _ { 2 } \right| = \left| z _ { 3 } \right| = 5,$  then  $z _ { 1 } \sin 2 A + z _ { 2 } \sin 2 B + z _ { 3 } \sin 2 C = 0.$

  1. True

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

This is a known identity in complex geometry for points on a circle centered at the origin. The sum of the vectors weighted by the sine of the angles relates to the geometry of the triangle inscribed in the circle.

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 $\sin \frac {6\pi}5+i(1+\cos \frac {6\pi }5)$ then

  1. $|Z|=-2\cos \frac {3\pi}5$
  2. $Arg(Z)=\frac {\pi}5$
  3. $Arg(Z)=\frac {9\pi }{10}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Z = sin(6pi/5) + i(1 + cos(6pi/5)). Using half-angle identities: sin(6pi/5) = 2sin(3pi/5)cos(3pi/5) and 1+cos(6pi/5) = 2cos^2(3pi/5). Factoring out 2cos(3pi/5) gives Z = 2cos(3pi/5) * (sin(3pi/5) + i cos(3pi/5)). The argument is 9pi/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

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.
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 $\overline { z } $ lies in the third quadrant then $z$ lies in the

  1. First quadrant

  2. Second quadrant

  3. Third quadrant

  4. Fourth quadrant
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Fact:  $\overline z$ is image of $z$ in $x$-axis

So if $\overline z$ lies in third quadrant then $z$ will lie in second quadrant 

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 $z _1$ and $z _2$ are two complex numbers such that $(1-i)z _1=2z _2$ and $arg(z _1z _2)=\dfrac{\pi}{2}$ then $arg(z _2)$ is equals to:

  1. $\dfrac{3 \pi}{8}$
  2. $\dfrac{\pi}{8}$
  3. $\dfrac{5 \pi}{8}$
  4. $\dfrac{-7 \pi}{8}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(1-i)z _1=2z _2$

$\cfrac{z _2}{z _1}=\cfrac{1}{2}-\cfrac{i}{2}$
Let $z _1=r _1e^{i\theta _1}\ and\ z _=r _1r^{\theta _2}$
$arg(\cfrac{z _2}{z _1})=\tan^{-1}\cfrac{-1/2}{1/2}=-\pi/4$
$\theta _2-\theta _1=-\pi/4$  and $arg(z _1z _2)=\pi/2$ (given)
$\implies \theta _2-\theta _1=-\pi/4$  and $\theta _2
+\theta _1=\pi/2$
$\implies 2\theta _2=\pi/4,\theta _2=\pi/8$
$\implies arg(z _2)=\pi/8$

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 complex number $\dfrac{1 + 2i}{1 - i}$ lies in which quadrant of the complex plane.

  1. First

  2. Second

  3. Third

  4. Fourth
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac{1+2i}{1-i}$
$\Rightarrow \dfrac{1+2i}{1-i}\times \dfrac{1+i}{1+i}$
$=\dfrac{1+i+2i-2i^2}{1-i^2}=\dfrac{1+3i-2}{2}$
$=\dfrac{-1+3i}{2}$
$\therefore$ It lies in $2^{nd}$ Quadrant.
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 $arg(z) < 0$, then $arg(-z)-arg(z)=$

  1. $\pi$
  2. $-\pi$
  3. $\dfrac{\pi}{2}$
  4. $-\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $Z=re^{i\theta _1}$

$-Z=-re^{i\theta _1}$
$\implies -a\cos\theta _1-ib\sin\theta _1$
$\implies -a\cos(\pi+\theta _1)-ib\sin(\pi+\theta _1)$
$\implies re^{i(\pi+\theta _1)}$
$arg(-Z)-arg(Z)$
$\implies \pi+\theta _1-\theta _1\ \implies \pi$

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

Which of the given alternatives represent a point in Argand plane, equidistant from roots of the equation $(z+1)^4= 16z^4$?

  1. $(0,0)$
  2. $\left(-\dfrac{1}{3},0\right)$
  3. $\left(\dfrac{1}{3},0\right)$
  4. $\left(0,\dfrac{2}{\sqrt5}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the given equation $(z+1)^4=16z^4$
$ \Rightarrow (z+1)^4=16z^4$

$ \Rightarrow z+1=(2^4z^4)^{\frac{1}{4}}$

$ \Rightarrow |z+1|=2|z|$

We know that $z=x+iy$

Therefore $|x+iy+1|=2|x+iy|$

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

$ \Rightarrow (x+1)^2+y^2=4(x^2+y^2)$

$ \Rightarrow x^2+2x+1+y^2=4x^2+4y^2$

$ \Rightarrow 3x^2+3y^2-2x-1=0$

Divide throughout by 3 we get,
$ \Rightarrow x^2+y^2-\dfrac{2}{3}x-\dfrac{1}{3}=0$, which represents a circle.

We know that for the circle equation of the form $x^2+y^2+2gx+2hy+c=0$ the center of the circle is given by $(-g,-h)$

We have $3x^2+3y^2-2x-1=0$ where $g=-\dfrac{1}{3}, h=0$.

Hence the center is $(\dfrac{1}{3},0)$ which is equidistant from the root of the 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

A particle starts from a point $z _0= I + i$, where $i
=\sqrt{-1}$ It moves horizontally away from origin by $2$ units and then
vertically away from origin by $3$ units to reach a point$ z _1$. From $z _1$
particle moves $\sqrt{5}$ units in the direction of $2\hat i + \hat j$ and
then it moves through an angle of $\cos e{c^{ - 1}}\sqrt 2 $ in anticlockwise
direction of a circle with centre at origin to reach a point $z _2$ . The arg $z _2$ is given by

  1. ${\sec ^{ - 1}}2$
  2. ${\cot ^{ - 1}}0$
  3. ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)$
  4. ${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The problem involves complex number transformations. Given the starting point and movements, the final argument calculation leads to an angle of 90 degrees, which corresponds to cot inverse of 0.