The value of |z|^2+|z-3|^2+|z-i|^2 is minimum | vedclass

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Question

(A) Let $z = x + iy$. Then $|z|^2 = x^2 + y^2$,$|z-3|^2 = (x-3)^2 + y^2$,and $|z-i|^2 = x^2 + (y-1)^2$. Let $f(x, y) = x^2 + y^2 + (x-3)^2 + y^2 + x^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $z = x + iy$. Then $|z|^2 = x^2 + y^2$,$|z-3|^2 = (x-3)^2 + y^2$,and $|z-i|^2 = x^2 + (y-1)^2$. Let $f(x, y) = x^2 + y^2 + (x-3)^2 + y^2 + x^2 + (y-1)^2$. $f(x, y) = 3x^2 - 6x + 9 + 3y^2 - 2y + 1 = 3(x^2 - 2x) + 3(y^2 - \frac{2}{3}y) + 10$. To minimize $f(x, y)$,we complete the square: $f(x, y) = 3(x-1)^2 - 3 + 3(y-\frac{1}{3})^2 - \frac{1}{3} + 10 = 3(x-1)^2 + 3(y-\frac{1}{3})^2 + \frac{20}{3}$. The function is minimized when $x = 1$ and $y = \frac{1}{3}$. Thus,$z = 1 + \frac{1}{3}i$.

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

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