Let z=x+iy and P(x, y) be a point on the Arga | vedclass

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(C) Let $z = x + iy$. The given condition is $\operatorname{Arg}\left(\frac{z-3i}{z+2i}\right) = \frac{\pi}{4}$.This represents the arc of a circle pa

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$x^2+y^2+5x-y-6=0, (x, y) 
eq (0, -2)$

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(C) Let $z = x + iy$. The given condition is $\operatorname{Arg}\left(\frac{z-3i}{z+2i}ight) = \frac{\pi}{4}$.This represents the arc of a circle passing through $A(0, 3)$ and $B(0, -2)$.Let $z_1 = 3i$ and $z_2 = -2i$. The expression $\frac{z-z_1}{z-z_2}$ has an argument of $\frac{\pi}{4}$.Using the property $\operatorname{Arg}(w) = 	heta \implies \operatorname{Im}(w) = 	an(	heta) \operatorname{Re}(w)$, we have $\frac{z-3i}{z+2i} = \frac{x+i(y-3)}{x+i(y+2)}$.Multiplying by the conjugate: $\frac{[x+i(y-3)][x-i(y+2)]}{x^2+(y+2)^2} = \frac{x^2 + (y-3)(y+2) + i[x(y-3) - x(y+2)]}{x^2+(y+2)^2}$.Thus, $\frac{x(y-3) - x(y+2)}{x^2 + (y-3)(y+2)} = 	an\left(\frac{\pi}{4}ight) = 1$.$-5x = x^2 + y^2 - y - 6$.Rearranging gives $x^2 + y^2 + 5x - y - 6 = 0$, where $(x, y) 
eq (0, -2)$.

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 $\|z\| \leq 3$

Let $z=x+iy$ and $P(x, y)$ be a point on the Argand plane. If $z$ satisfies the condition $\operatorname{Arg}\left(\frac{z-3i}{z+2i}\right)=\frac{\pi}{4}$, then the locus of $P$ is:

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