Tag: argand plane and polar representation

Questions Related to argand plane and polar representation

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. $4q^{2}\ \cos^{2} \left(\dfrac{\alpha}{2}\right)$


Correct Option: A

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


Correct Option: A

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


Correct Option: C

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


Correct Option: A
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.

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


Correct Option: B
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 

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. $\dfrac{-7 \pi}{8}$


Correct Option: B
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 $\theta _2
+\theta _1=\pi/2$

$\implies 2\theta _2=\pi/4,\theta _2=\pi/8$
$\implies arg(z _2)=\pi/8$

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

  1. First

  2. Second

  3. Third

  4. Fourth


Correct Option: B
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.

If $arg(z) < 0$, then $arg(-z)-arg(z)=$

  1. $\pi$

  2. $-\pi$

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

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


Correct Option: A
Explanation:

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

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

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. $\left(0,\dfrac{2}{\sqrt5}\right)$


Correct Option: C
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.

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. ${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$


Correct Option: B