If the amplitude of (z-1-2i) is frac{pi}{3},t | vedclass

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(A) Let $z = x + iy$. Then $z - 1 - 2i = (x - 1) + i(y - 2)$.Given that $	ext{arg}(z - 1 - 2i) = \frac{\pi}{3}$.This implies $	an\left(\frac{\pi}{3}

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$y=\sqrt{3}x+(2-\sqrt{3})$

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(A) Let $z = x + iy$. Then $z - 1 - 2i = (x - 1) + i(y - 2)$.Given that $	ext{arg}(z - 1 - 2i) = \frac{\pi}{3}$.This implies $	an\left(\frac{\pi}{3}ight) = \frac{y - 2}{x - 1}$,where $x > 1$ and $y > 2$.Since $	an\left(\frac{\pi}{3}ight) = \sqrt{3}$,we have $\sqrt{3} = \frac{y - 2}{x - 1}$.Rearranging the terms,we get $y - 2 = \sqrt{3}(x - 1)$.$y = \sqrt{3}x - \sqrt{3} + 2$.$y = \sqrt{3}x + (2 - \sqrt{3})$.

If the amplitude of $(z-1-2i)$ is $\frac{\pi}{3}$,then the locus of $z$ is

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