A(z_1=2+2i),B(z_2),and C(z_3) are three point | vedclass

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(C) The equation $|z-2i|=2$ represents a circle in the Argand plane with center at $(0, 2)$ and radius $r=2$.For $	riangle ABC$ to have the maximum a

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$4$

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(C) The equation $|z-2i|=2$ represents a circle in the Argand plane with center at $(0, 2)$ and radius $r=2$.For $	riangle ABC$ to have the maximum area,it must be an equilateral triangle inscribed in the circle.Let $A$ be the point $(2, 2)$. The center of the circle is $O'(0, 2)$.The line segment $AO'$ lies on the horizontal line $y=2$.For an equilateral triangle,the altitude from $A$ to the side $BC$ must pass through the center $O'(0, 2)$.Since $A$ is at $(2, 2)$ and $O'$ is at $(0, 2)$,the altitude $AM$ lies on the line $y=2$.Thus,the side $BC$ is a vertical chord passing through $M(-2, 2)$.Since $BC$ is a vertical line,the $x$-coordinate of both $B$ and $C$ is $-2$.The points $B$ and $C$ lie on the circle $x^2 + (y-2)^2 = 4$.Substituting $x=-2$: $(-2)^2 + (y-2)^2 = 4 \implies 4 + (y-2)^2 = 4 \implies (y-2)^2 = 0 \implies y=2$.Wait,if $y=2$,then $B$ and $C$ coincide at $(-2, 2)$,which is not possible for a triangle.Re-evaluating: The altitude from $A(2, 2)$ to $BC$ is the line segment $AM$. Since the circle is centered at $(0, 2)$,the altitude must be the diameter along the $x$-axis direction. The point $M$ is the midpoint of $BC$. The coordinates of $M$ are $(-1, 2)$ because $A$ is at $(2, 2)$ and the center is $(0, 2)$,the distance $AO'=2$. For an equilateral triangle,the distance from the center to the side is $r/2 = 1$. Thus $M$ is at $2-3= -1$ on the $x$-axis.Therefore,the $y$-coordinate of $M$ is $2$. Since $M$ is the midpoint of $BC$,$\frac{	ext{Im}(z_2) + 	ext{Im}(z_3)}{2} = 	ext{Im}(M) = 2$.Thus,$	ext{Im}(z_2) + 	ext{Im}(z_3) = 2 	imes 2 = 4$.

$A(z_1=2+2i)$,$B(z_2)$,and $C(z_3)$ are three points on the Argand plane satisfying $|z_k-2i|=2$ for $k=1, 2, 3$. If $\triangle ABC$ encloses the maximum area,then the sum of the imaginary parts of $z_2$ and $z_3$ is

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