If a complex number z is such that frac{z-2i} | vedclass

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(A) Let $z = x + iy$. The condition that $\frac{z-2i}{z-2}$ is purely imaginary implies that its real part is $0$. Alternatively,$\frac{z-2i}{z-2} + \

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$2\pi$

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(A) Let $z = x + iy$. The condition that $\frac{z-2i}{z-2}$ is purely imaginary implies that its real part is $0$. Alternatively,$\frac{z-2i}{z-2} + \overline{\left(\frac{z-2i}{z-2}\right)} = 0$. $\frac{z-2i}{z-2} + \frac{\bar{z}+2i}{\bar{z}-2} = 0$. $(z-2i)(\bar{z}-2) + (\bar{z}+2i)(z-2) = 0$. $z\bar{z} - 2z - 2i\bar{z} + 4i + z\bar{z} - 2\bar{z} + 2iz - 4i = 0$. $2|z|^2 - 2(z+\bar{z}) + 2i(z-\bar{z}) = 0$. $|z|^2 - (z+\bar{z}) + i(z-\bar{z}) = 0$. Substituting $z = x+iy$,we get $x^2 + y^2 - 2x - 2y = 0$. $(x-1)^2 + (y-1)^2 = 2$. This represents a circle with center $(1, 1)$ and radius $r = \sqrt{2}$. The circle passes through the origin $(0,0)$. The entire circle lies in the first quadrant. Area $= \pi r^2 = \pi(\sqrt{2})^2 = 2\pi$.

If a complex number $z$ is such that $\frac{z-2i}{z-2}$ is a purely imaginary number and the locus of $z$ is a closed curve,then the area of the region bounded by that closed curve and lying in the first quadrant is

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