Question

State true or false for the following:
The order relation is defined on the set of complex numbers.

State true or false for the following:
Multiplication of a non-zero complex number by $-i$ rotates the point about origin through a right angle in the anti-clockwise direction.

State true or false for the following:
For any complex number $z$, the minimum value of $|z|+|z-1|$ is $1$.

State true or false for the following:
The locus represented by $|z-1|=|z-i|$ is a line perpendicular to the line joining points $(1,0)$ and $(0,1)$.

State true or false for the following:
If $z$ is a complex number such that $z\ne 0$ and $Re\left(z\right)=0$, then $Im\left(z^2\right)=0$.

State true or false for the following:
The inequality $|z-4|<|z-2|$ represents the region given by $x>3$.

State true or false for the following:
Let $z_1$ and $z_2$ be two complex numbers such that $|z_1+z_2|=|z_1|+|z_2|$, then arg⁡ $(z_1-z_2)=0$.

State true or false for the following:
$2$ is not a complex number.

Answer 1

We can compare two complex numbers when they are purely real.
Otherwise, comparison of complex numbers has no meaning.
Hence, the given statement is false.


Method 1: Algebraic Method

Assuming a non-zero complex
number as $z=|z|(\cos \theta +i\sin \theta )$

Multiplying $z$ by $-i$, we get

$-iz=-i|z|(\cos \theta +i\sin \theta )$

$\Rightarrow -iz=|z|(-i\cos \theta -i^2\sin \theta )$

$\Rightarrow -iz=|z|(\sin \theta -i\cos \theta )$ $[\because i^2=-1]$

$\Rightarrow -iz=|z|(\cos (\frac{\pi }{2}-\theta )-i\sin (\frac{\pi }{2}-\theta ))$

$\Rightarrow -iz=|z|(\cos (\theta -\frac{\pi }{2})+i\sin (\theta -\frac{\pi }{2}))$

Therefore, arg⁡ $\left(z\right)=\theta$ and
arg⁡ $(-iz)=\theta -\frac{\pi }{2}$,

so $z$ is rotated by $\frac{\pi }{2}$ but in clockwise direction.

Hence, the given statement is false.

Method 2 : Geometrical Method

Assuming a non-zero complex
number as $z=x+iy$

Taking $x,y>0$

Multiplying $z$ by $-i$, we get

$-iz=-i(x+iy)=-ix-i^{2}y$

$\Rightarrow -iz=y-ix$ $[\because i^2=-1]$

Plotting $(x,y)$ and $(y,-x)$ on Argand plane.


Now, $OA=\sqrt{x^2+y^2},OB=\sqrt{y^2+x^2},$

$AB=\sqrt{(x-y{)}^2+(y+x{)}^2}=\sqrt{2x^2+2y^2}$

We can see that,

$O{A}^2+O{B}^2=A{B}^2$

Therefore, $\angle AOB=\frac{\pi }{2}$, but the rotation is in clockwise direction.

Hence, the given statement is false.


We know that,

$|a-b|\le |a|+|b|$

Taking $a=z,b=z-1$

$\Rightarrow |z-(z-1)|\le |z|+|z-1|$

$\Rightarrow |z|+|z-1|\ge 1$

So, the minimum value of

$|z|+|z-1|$ is $1$.

Hence, the given statement is True.


Finding the required locus.

Given : $|z-1|=|z-i|$

Let $z=x+iy$, then

$|x+iy-1|=|x+iy-i|$

$\Rightarrow |(x-1)+iy|=|x+i(y-1)|$

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

Squaring on both sides, we get

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

$\Rightarrow y=x$

Locus is a line, whose slope is $m_1=1$

Checking the perpendicularity of the lines.

Slope of the line joining points $(1,0)$ and $(0,1)$

$m_2=\frac{0-1}{1-0}=-1$

As, $m_1\cdot m_2=-1$, so both lines are perpendicular.

Hence, the given statement is True.


Given:

$z\ne 0$ and $Re\left(z\right)=0$

So, $z=iy,y\ne 0$​

Now, $z^2=(iy{)}^2=i^2{y}^2=-y^2$​

$[\because i^2=-1]$

Therefore, $Im\left(z^2\right)=0$​​​​​

Hence, the given statement is True.


Given:

$|z-4|<|z-2|$

Let $z=x+iy$

$\Rightarrow |x-4+iy|<|x-2+iy|$

$\Rightarrow \sqrt{(x-4{)}^2+y^2}<\sqrt{(x-2{)}^2+y^2}$

Squaring on both sides, we get

$\Rightarrow (x-4{)}^2+y^2<(x-2{)}^2+y^2$

$\Rightarrow x^2+16-8x<x^2+4-4x$

$\Rightarrow 4x>12$

$\therefore x>3$

Hence, the above statement is true.


Given:

$|z_1+z_2|=|z_1|+|z_2|$

Squaring on both sides, we get

$\Rightarrow |z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2+2|z_1||z_2|$

$\Rightarrow |z_1{|}^2+|z_2{|}^2+2Re\left(z_1\overline{z_2}\right)=|z_1{|}^2+|z_2{|}^2+2|z_1||z_2|$

$\{\because |z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2+2Re(z_1\overline{z_2}\left)\right\}$

$\Rightarrow 2Re\left(z_1\overline{z_2}\right)=2|z_1||z_2|$

$\Rightarrow Re\left(z_1\overline{z_2}\right)=|z_1||z_2|$ ...$\left(1\right)$

Assuming $z_1=|z_1|(\cos \alpha +i\sin \alpha )$,

$z_2=|z_2|(\cos \beta +i\sin \beta )$

$z_1\overline{z_2}=|z_1||z_2|(\cos \alpha +i\sin \alpha )(\cos \beta -i\sin \beta )$

$\Rightarrow z_1\overline{z_2}=|z_1||z_2|\times \left[\begin{array}{c}\cos \alpha \cos \beta -i^2\sin \alpha \sin \beta \\ +i(\sin \alpha \cos \beta -\cos \alpha \sin \beta )\end{array}\right]$

$\Rightarrow z_1\overline{z_2}=|z_1||z_2|\left[\begin{array}{c}\cos (\alpha -\beta )+\\ i\sin (\alpha -\beta )\end{array}\right]$

$[\because i^2=-1]$

From equation $\left(1\right)$, we get

$\Rightarrow |z_1||z_2|\cos (\alpha -\beta )=|z_1||z_2|$

$\Rightarrow \cos (\alpha -\beta )=1$

$\Rightarrow \alpha -\beta =0$

$\Rightarrow \alpha =\beta$

Now,

$z_1-z_2=|z_1|(\cos \alpha +i\sin \alpha )-|z_2|(\cos \beta +i\sin \beta )$

$\Rightarrow z_1-z_2=(|z_1|-|z_2|)(\cos \alpha +i\sin \alpha )$

$[\because \alpha =\beta ]$

Therefore, arg $(z_1-z_2)=\alpha =\beta ,$

which is not always zero.

Hence, the given statement is False.


Any general complex number is of

form $z=x+iy,$ $x,y\in R$

$2$ is a complex number because

$x=2,y=0$ which satisfy the definition of complex numbers.

Hence, the given statement is False.