Describe the set of points z=x+iy satisfying |z-1|>2|z-i|
Describe the set of points $z=x+iy$ satisfying $\ |z-1|>2|z-i|$.
Explanation of Method
Square both sides (valid since both are nonnegative), substitute $z=x+iy$, expand, and complete the square in $x$ and $y$ to recognize the boundary as a circle — an instance of the Apollonius-circle phenomenon for ratio-of-distance conditions with ratio $\neq1$.
Step 1 · Square both sides
$$ |z-1|^2>4|z-i|^2 $$
Step 2 · Substitute $z=x+iy$
$$ (x-1)^2+y^2>4\left[x^2+(y-1)^2\right] $$
Step 3 · Expand both sides
$$ x^2-2x+1+y^2>4x^2+4y^2-8y+4 $$
Step 4 · Collect all terms on one side
$$ 0>3x^2+3y^2+2x-8y+3 \quad\Longrightarrow\quad 3x^2+3y^2+2x-8y+3<0 $$
Dividing by $3$:
$$ x^2+y^2+\frac23x-\frac83y+1<0 $$
Step 5 · Complete the square
$$ \left(x+\frac13\right)^2-\frac19+\left(y-\frac43\right)^2-\frac{16}9+1<0 $$
$$ \left(x+\frac13\right)^2+\left(y-\frac43\right)^2<\frac19+\frac{16}9-1=\frac{17}9-\frac99=\frac89 $$
Step 6 · Conclusion
$$ \boxed{\left(x+\frac13\right)^2+\left(y-\frac43\right)^2<\frac89} $$
an open disk centered at $z_0=-\dfrac13+\dfrac43i$ with radius $r=\sqrt{\dfrac89}=\dfrac{2\sqrt2}3$.
The open disk of radius $\frac{2\sqrt2}3$ centered at $-\frac13+\frac43i$ — points strictly closer to $1$ (in the sense of the $2\!:\!1$ ratio) lie inside.
Verification
Test point $z_0=-\frac13+\frac43i$ (the center): $|z_0-1|=\left|-\frac43+\frac43i\right|=\frac{4\sqrt2}3$, and $2|z_0-i|=2\left|-\frac13+\frac13i\right|=\frac{2\sqrt2}3$. Indeed $\frac{4\sqrt2}3>\frac{2\sqrt2}3$, confirming the center satisfies the original inequality. The algebraic simplification was also checked symbolically term by term.
Result
A distance-ratio condition $|z-a|=k|z-b|$ with $k\neq1$ always defines a circle (the Apollonius circle) rather than a line; here $k=2$ produces the open disk bounded by that circle.