Let A, B, C be three sets of complex numbers | vedclass

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(C) Let $z = x + iy$. The set $A$ is $y \ge 1$.For set $B$,$|(x-2) + i(y-1)| = 3$,so $(x-2)^2 + (y-1)^2 = 9$.For set $C$,$	ext{Re}((1-i)(x+iy)) = 	e

Vedclass · Accepted Answer

$35$ and $39$

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(C) Let $z = x + iy$. The set $A$ is $y \ge 1$.For set $B$,$|(x-2) + i(y-1)| = 3$,so $(x-2)^2 + (y-1)^2 = 9$.For set $C$,$	ext{Re}((1-i)(x+iy)) = 	ext{Re}(x + iy - ix + y) = x + y = \sqrt{2}$.Substituting $y = \sqrt{2} - x$ into the equation for $B$: $(x-2)^2 + (\sqrt{2} - x - 1)^2 = 9$.$(x-2)^2 + (x - (\sqrt{2}-1))^2 = 9$.Solving this quadratic equation for $x$ and checking the condition $y \ge 1$ shows that the intersection $A \cap B \cap C$ is a single point.Using the property of the sum of squares of distances from two points $z_1 = -1+i$ and $z_2 = 5+i$ to a point $z$ on the circle,the expression $|z - z_1|^2 + |z - z_2|^2$ is constant if $z$ is the midpoint of the diameter.Here,the distance between $z_1$ and $z_2$ is $|(5+i) - (-1+i)| = |6| = 6$.Since $z$ lies on the circle with diameter $6$,the sum of the squares of the distances to the endpoints of the diameter is $(6)^2 = 36$.Thus,the value is $36$,which lies between $35$ and $39$.

Let $A, B, C$ be three sets of complex numbers defined as $A = \{z : \text{Im}(z) \ge 1\}$,$B = \{z : |z - 2 - i| = 3\}$,and $C = \{z : \text{Re}((1 - i)z) = \sqrt{2}\}$. If $z$ is any point in $A \cap B \cap C$,then $|z + 1 - i|^2 + |z - 5 - i|^2$ lies between:

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