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State True of False for the following:

i) The order relation is defined on the set of complex numbers.

ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction

iii) For any complex number z, the minimum value of |z| + |z – 11 is 1

iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1)

v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z²) = 0

vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z₁ and Z₂ be two complex numbers such that |z, + z₂| = |z₁ j + |z₂|, then arg (z₁ – z₂) = 0.
(viii) 2 is not a complex number.

Answers (1)

(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex number can be compared. So, it is false.

(ii) Let $z= x+iy$

$z.i = (x+iy)i =xi= -y$ which rotates at angle of 180. So, it is ‘false’.

(iii) Let $z= x+iy$

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

The value of $|z|+|z-1|$ is minimum when $x=0,y=0 \, \: \: i.e.,1$

Hence, it is true.

iv) Let $z= x+iy$

Given that $\left | z-1 \right |=\left |z-i \right |$

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

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

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

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

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

$-2x+2y=0 x-y=0$ which is a straight line slope=1

Now, equation of line through the point 1,0and 0,1

$y-0=\frac{1-0}{0-1} (x-1)$

$y=-x+1$ whose slope=-1

Multiplication of the slopes of two lines =-1*1=-1

So, they are perpendicular. Hence, it is true.

v)Let $z= x+iy$ $z\neq0 \: \: and \: \: Re(z)=0$

Since, real part is 0 $x=0, z=0+iy=iy$

$1m(z^2 )=y^2 i^2=-y^2$ which is real Hence, it is False.

vi) $(z-4)<|z-2|$

Let $z= x+iy$

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

$|(x-4)+iy|<|(x-2)+iy|$

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

$(x-4)^2+y^2<(x-2)^2+y^2$

$(x-4)^2<(x-2)$

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

$8x+4x<-16+4$

$-4x<-12$

$x>3$ Hence, it is true.

vii) Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$

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

$|x_1+iy_1+x_2+iy_2 |=|(x_1+iy_1 )+(x_2+y_2 i)|$

$= \sqrt{(x_1+x_2 )^2+(y_1+y_2 )^2 }$

$=\sqrt{x_1^2+y_1^2} +\sqrt{x_2^2+y_2^2 }$

Squaring both sides, we get $(x_1^2+x_2^2+2x_1 x_2+y_1^2+y_2^2+2y_1 y_2$ )

$=x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

$=2x_1 x_2+2y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

$=x_1 x_2+y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

Again squaring on both sides we get $x_1^2 x_2^2+x_1^2 y_2^2+2x_1 y_1 x_2 y_2=x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 2x_1 y_1 x_2 y_2=x_1^2 y_2^2+x_2^2 y_1^2$

$x_1^2 y_1^2+x_2^2 y_1^2-2x_1 y_1 x_2 y_2=0$

$(x_1 y_2-x_2 y_1 )^2=0$

$x_1 y_2-x_2 y_1=0$

$x_1 y_2=x_2 y_1$

$\frac{x_1}{y_1}=\frac{x_2}{y_2}$

$\frac{y_1}{x_1}=\frac{y_2}{x_2}$

$arg(z_1)=arg(z_2)=0$ Hence, it is false.

(viii) Since, every real number is a complex number. So, 2 is a complex number. Hence, it is false.

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