Let S=z=x+iy: |z-1+i||z|, |z|

Question

Let $S=\{z=x+iy: |z-1+i|\ge|z|,\ |z|<2,\ |z+i|=|z-1|\}$. Then the set of all values of $x$, for which $u=2x+iy\in S$ for some $y\in\mathbb R$, is

Accepted Answer

From $|z+i|=|z-1|$ (equidistant from $(0,-1)$ and $(1,0)$): $2y=-2x \implies y=-x$.

From $|z|<2$: $x^2+y^2<4 \implies 2x^2<4 \implies -\sqrt2<x<\sqrt2$.

From $|z-1+i|\ge|z|$ with $y=-x$: $(x-1)^2+(1-x)^2 \ge x^2+x^2 \implies 2(x-1)^2\ge2x^2 \implies x^2-2x+1\ge x^2 \implies x\le\dfrac12$.

So for $z=x+iy\in S$: $x\in\left(-\sqrt2,\dfrac12\right]$.

For $u=2x+iy\in S$, replace $x$ by $2x$ in this range: $-\sqrt2<2x\le\dfrac12 \implies x\in\left(-\dfrac1{\sqrt2},\dfrac14\right]$.

Solutions · 1