The value of |z|^{2}+|z-3|^{2}+|z-i|^{2} is m | vedclass

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(C) Let $z = x + iy$.Then the expression becomes $|x+iy|^2 + |(x-3)+iy|^2 + |x+i(y-1)|^2$.$= (x^2 + y^2) + ((x-3)^2 + y^2) + (x^2 + (y-1)^2)$.$= x^2 +

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$1+\frac{i}{3}$

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(C) Let $z = x + iy$.Then the expression becomes $|x+iy|^2 + |(x-3)+iy|^2 + |x+i(y-1)|^2$.$= (x^2 + y^2) + ((x-3)^2 + y^2) + (x^2 + (y-1)^2)$.$= x^2 + y^2 + x^2 - 6x + 9 + y^2 + x^2 + y^2 - 2y + 1$.$= 3x^2 - 6x + 3y^2 - 2y + 10$.$= 3(x^2 - 2x + 1) + 3(y^2 - \frac{2}{3}y + \frac{1}{9}) + 10 - 3 - \frac{1}{3}$.$= 3(x-1)^2 + 3(y - \frac{1}{3})^2 + \frac{20}{3}$.This expression is minimum when $x-1 = 0$ and $y - \frac{1}{3} = 0$.Therefore, $x = 1$ and $y = \frac{1}{3}$.Thus, $z = 1 + \frac{i}{3}$.

The value of $|z|^{2}+|z-3|^{2}+|z-i|^{2}$ is minimum when $z$ equals

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