(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 the origin through a right angle in the anti-clockwise direction.

(iii) For any complex number $z$ , the minimum value of $?z+\frac{1}{z}?$ is 1.

(iv) The locus represented by $?z-1?=?z-i?$ is a line perpendicular to the join of $\left(1,0\right)$ and $\left(0,1\right)$ .

(v) If $z$ is a complex number such that $z\ne 0$ and Re$\left(z\right)=0$ , then Im$\left(z^2\right)=0$ .

(vi) The inequality $?z-4?<?z-2?$ represents the region given by $x>3$ .

(vii) Let $z_1$ and $z_2$ be two complex numbers such that $?z_1+z_2?=?z_1?+?z_2?$ , then arg$\left(z_1-z_2\right)=0$ .

(viii) $2$ is not a complex number.$\mathrm{}$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}\left(i\right)Comparisonoftwopurelyimaginarycomplexnumbersisnotpossible.However,thetwo\\ purelyimaginaryrealcomplexnumberscanbecompared.\\ So,itis\text{'}False\text{'}.\\ \left(ii\right)Letz=x+yi\\ z.i=\left(x+yi\right)i=xi-ywhichrotatesatangleof180^0\\ So,itis\text{'}False\text{'}.\\ \left(iii\right)Letz=x+yi\\ \therefore \left|z\right|+\left|z+1\right|=\sqrt[]{x^2+y^2}+\sqrt[]{{\left(x-1\right)}^2+y^2}\\ Thevalueof\left|z\right|+\left|z+1\right|isminimumwhenx=0,y=0i.e.,1.\\ Hence,itis\text{'}True\text{'}.\\ \left(iv\right)Letz=x+yi\\ Giventhat:\left|z-1\right|=\left|z-i\right|\\ then\left|x+yi-1\right|=\left|x+yi-i\right|\\ \Rightarrow \left|\left(x-1\right)+yi\right|=\left|x-\left(1-y\right)i\right|\\ \Rightarrow \sqrt[]{{\left(x-1\right)}^2+y^2}=\sqrt[]{x^2+{\left(1-y\right)}^2}\\ \Rightarrow {\left(x-1\right)}^2+y^2=x^2+{\left(1-y\right)}^2\\ \Rightarrow x^2-2x+1+y^2=x^2+1+y^2-2y\\ \Rightarrow -2x+2y=0\\ \Rightarrow x-y=0whichisastraightline.\\ Slope=1\\ Nowequationofalinethroughthepoint\left(1,0\right)and\left(0,1\right)\\ y-0=\frac{1-0}{0-1}\left(x-1\right)\\ \Rightarrow y=-x+1whoseslope=-1.\\ Nowthemultiplicationoftheslopesoftwolines=-1\times 1=-1,\\ Sotheyareperpendicular.\\ Hence,itis\text{'}True\text{'}.\end{array}$