If z=x+iy,where x, y in mathbb{R} and the poi | vedclass

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(A) Given $\arg \left(\frac{z-1}{z-3i}\right)=\frac{\pi}{2}$.Let $z=x+iy$. Then $\frac{z-1}{z-3i} = \frac{(x-1)+iy}{x+i(y-3)}$.Multiplying numerator a

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$\left\{z \in \mathbb{C} : \left|z-\frac{1+3i}{2}\right|=\frac{\sqrt{10}}{2}\right\}$

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(A) Given $\arg \left(\frac{z-1}{z-3i}\right)=\frac{\pi}{2}$.Let $z=x+iy$. Then $\frac{z-1}{z-3i} = \frac{(x-1)+iy}{x+i(y-3)}$.Multiplying numerator and denominator by the conjugate of the denominator: $\frac{((x-1)+iy)(x-i(y-3))}{x^2+(y-3)^2} = \frac{x(x-1)+y(y-3) + i(xy - (x-1)(y-3))}{x^2+(y-3)^2}$.For the argument to be $\frac{\pi}{2}$,the real part must be $0$ and the imaginary part must be positive.Real part: $x(x-1)+y(y-3)=0 \Rightarrow x^2-x+y^2-3y=0$.Completing the square: $\left(x-\frac{1}{2}\right)^2 + \left(y-\frac{3}{2}\right)^2 = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \left(\frac{\sqrt{10}}{2}\right)^2$.This represents a circle with center $\left(\frac{1}{2}, \frac{3}{2}\right)$ and radius $\frac{\sqrt{10}}{2}$.The condition $\arg(w) = \frac{\pi}{2}$ implies the locus is the arc of the circle where the imaginary part is positive.

If $z=x+iy$,where $x, y \in \mathbb{R}$ and the point $P$ in the Argand plane represents $z$,then the locus of $P$ satisfying the condition $\arg \left(\frac{z-1}{z-3i}\right)=\frac{\pi}{2}$ is:

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