The sum of the square of the modulus of the e | vedclass

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(B) Given $|z-1| \leq 1$,where $z=x+iy$. $(x-1)^2 + y^2 \leq 1$. This represents a disk centered at $(1,0)$ with radius $1$. Also,$|z-5| \leq |z-5i|$.

Vedclass · Accepted Answer

$9$

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(B) Given $|z-1| \leq 1$,where $z=x+iy$. $(x-1)^2 + y^2 \leq 1$. This represents a disk centered at $(1,0)$ with radius $1$. Also,$|z-5| \leq |z-5i|$. $(x-5)^2 + y^2 \leq x^2 + (y-5)^2$. $x^2 - 10x + 25 + y^2 \leq x^2 + y^2 - 10y + 25$. $-10x \leq -10y \Rightarrow x \geq y$. We need to find $z=x+iy$ such that $x, y \in \mathbb{Z}$ satisfying $(x-1)^2 + y^2 \leq 1$ and $x \geq y$. Possible integer points $(x,y)$ satisfying $(x-1)^2 + y^2 \leq 1$: If $x=0$,$(0-1)^2 + y^2 \leq 1$ $\Rightarrow 1 + y^2 \leq 1$ $\Rightarrow y^2 \leq 0$ $\Rightarrow y=0$. Point: $(0,0)$. Check $x \geq y$: $0 \geq 0$ (True). If $x=1$,$(1-1)^2 + y^2 \leq 1$ $\Rightarrow y^2 \leq 1$ $\Rightarrow y \in \{-1, 0, 1\}$. Points: $(1,-1), (1,0), (1,1)$. Check $x \geq y$: $1 \geq -1$ (True),$1 \geq 0$ (True),$1 \geq 1$ (True). If $x=2$,$(2-1)^2 + y^2 \leq 1$ $\Rightarrow 1 + y^2 \leq 1$ $\Rightarrow y^2 \leq 0$ $\Rightarrow y=0$. Point: $(2,0)$. Check $x \geq y$: $2 \geq 0$ (True). The set of elements is $z \in \{0, 1-i, 1, 1+i, 2\}$. The sum of the squares of the modulus is: $|0|^2 + |1-i|^2 + |1|^2 + |1+i|^2 + |2|^2 = 0 + (1^2+(-1)^2) + 1^2 + (1^2+1^2) + 2^2 = 0 + 2 + 1 + 2 + 4 = 9$.

The sum of the square of the modulus of the elements in the set $\{z=a+ib: a, b \in \mathbb{Z}, z \in \mathbb{C}, |z-1| \leq 1, |z-5| \leq |z-5i|\}$ is ........

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