Match the statements in column-I with those i | vedclass

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(C) $(A)-(q)$: $|z-i|z||=|z+i|z|| \Rightarrow |\frac{z}{|z|}-i|=|\frac{z}{|z|}+i|$,for $z 
eq 0$. The expression $\frac{z}{|z|}$ represents a point o

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$A-q, B-p, C-p, t, D-q, t$

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(C) $(A)-(q)$: $|z-i|z||=|z+i|z|| \Rightarrow |\frac{z}{|z|}-i|=|\frac{z}{|z|}+i|$,for $z 
eq 0$. The expression $\frac{z}{|z|}$ represents a point on the unit circle. The equation implies that the point $\frac{z}{|z|}$ is equidistant from $i$ and $-i$. The locus of such points is the real axis,where $\operatorname{Im}(z)=0$.$(B)-(p)$: $|z+4|+|z-4|=10$ represents an ellipse with foci at $(\pm 4, 0)$ and major axis length $2a=10$. Here $2ae=8$ and $2a=10$,so $e=4/5$. Thus,it is an ellipse with eccentricity $4/5$.$(C)-(p), (t)$: Let $\omega=2(\cos 	heta+i \sin 	heta)$. Then $z = 2(\cos 	heta+i \sin 	heta) - \frac{1}{2}(\cos 	heta-i \sin 	heta) = \frac{3}{2} \cos 	heta + i \frac{5}{2} \sin 	heta$. This is an ellipse $\frac{x^2}{(3/2)^2} + \frac{y^2}{(5/2)^2} = 1$ with $e^2 = 1 - \frac{9/4}{25/4} = 16/25$,so $e=4/5$. Since the semi-major axis is $2.5 $(D)-(q), (t)$: Let $\omega = \cos 	heta + i \sin 	heta$. Then $z = (\cos 	heta + i \sin 	heta) + (\cos 	heta - i \sin 	heta) = 2 \cos 	heta$. Since $z$ is purely real,$\operatorname{Im}(z)=0$ and $|z| = |2 \cos 	heta| \leq 2 < 3$.

column-$I$	column-$II$
$(A)$ The set of points $z$ satisfying $\|z-i\|z\|\|=\|z+i\|z\|\|$ is contained in or equal to	$(p)$ an ellipse with eccentricity $\frac{4}{5}$
$(B)$ The set of points $z$ satisfying $\|z+4\|+\|z-4\|=10$ is contained in or equal to	$(q)$ the set of points $z$ satisfying $\operatorname{Im} z=0$
$(C)$ If $\|\omega\|=2$,then the set of points $z=\omega-1/\omega$ is contained in or equal to	$(r)$ the set of points $z$ satisfying $\|\operatorname{Im} z\| \leq 1$
$(D)$ If $\|\omega\|=1$,then the set of points $z=\omega+1/\omega$ is contained in or equal to	$(s)$ the set of points $z$ satisfying $\|\operatorname{Re} z\| \leq 1$
	$(t)$ the set of points $z$ satisfying $\|z\| \leq 3$

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