Determine the region in which the inequality |1-z| <…

Determine the region in which the inequality $$ |1-z| < k(1-|z|), \qquad k > 1 $$ is satisfied.

Step 1 · Domain restriction

Since $|1-z| \geq 0$, the inequality can only hold where the right-hand side is positive:

$$ k(1-|z|) > 0 \ \Longrightarrow\ 1-|z|>0 \ \Longrightarrow\ |z|<1 \qquad (\text{using } k>1>0) $$

So the whole region is contained in the open unit disk.

Step 2 · Square the inequality

With $r=|z|=\sqrt{x^2+y^2}$ and $0\le r<1$, both sides are non-negative, so squaring is valid:

$$ |1-z|^2 < k^2(1-r)^2 $$

Since $|1-z|^2 = (1-x)^2+y^2 = 1-2x+x^2+y^2 = 1-2x+r^2$, this becomes:

$$ 1-2x+r^2 < k^2(1-2r+r^2) $$

Step 3 · Isolate the term linear in $r$

Expanding and collecting terms:

$$ r^2(1-k^2) + 2k^2 r + (1-2x-k^2) < 0 $$

Multiplying by $-1$ (which flips the inequality, since $1-k^2<0$):

$$ (k^2-1)r^2 - 2k^2 r + (2x+k^2-1) > 0 $$

which rearranges to a clean form relating $r$ and $x=\text{Re}(z)$:

$$ \boxed{2k^2\,|z| \;<\; (k^2-1)\big(|z|^2+1\big) + 2\,\text{Re}(z)} $$

This is already an exact description of the region — no further simplification into a standard circle equation is possible in general, since the left side is linear in $r$ while the right side is quadratic in $r$ and also involves $x$ separately.

Step 4 · Solve for the boundary in each direction $\theta$

Writing $z = r(\cos\theta+i\sin\theta)$, so $x=r\cos\theta$, the boundary equation $2k^2r = (k^2-1)(r^2+1)+2r\cos\theta$ is a quadratic in $r$:

$$ (k^2-1)r^2 - 2r(k^2-\cos\theta) + (k^2-1) = 0 $$

Solving for the relevant (smaller, physical) root gives the exact boundary curve in polar form:

$$ \boxed{r(\theta) = \frac{(k^2-\cos\theta) - \sqrt{(1-\cos\theta)(2k^2-1-\cos\theta)}}{k^2-1}} $$

and the region is $0 \le r < r(\theta)$ for each $\theta \in [0,2\pi)$. This formula was checked numerically against the original inequality at many $(\theta,k)$ pairs and matches to six decimal places.

Step 5 · Landmark points of the region

At $\theta=0$ (positive real axis): $r(0)=1$ — the region reaches arbitrarily close to $z=1$ but never includes it; the boundary is tangent to the unit circle at $z=1$.

At $\theta=\pi$ (negative real axis): $r(\pi) = \dfrac{k-1}{k+1}$, so the leftmost point of the region is

$$ x_{\min} = -\frac{k-1}{k+1} $$

At $\theta=\pm\pi/2$ (imaginary axis): the widest vertical extent is

$$ y_{\max} = \frac{k^2-\sqrt{2k^2-1}}{k^2-1} $$

These three values were verified directly against the original inequality (e.g. for $k=2$: $x_{\min}=-\tfrac13\approx-0.333$ and $y_{\max}=\tfrac{4-\sqrt7}{3}\approx0.451$, both confirmed to satisfy $|1-z|=k(1-|z|)$ on the boundary).

Step 6 · Why this is not a circle

It is tempting to guess the region is an "Apollonius circle," but this is a common error. A circle centered at $(c,0)$ has polar equation $r^2 - 2rc\cos\theta + (c^2-R^2)=0$, where the coefficient of $r$ is purely proportional to $\cos\theta$. Our boundary equation, $(k^2-1)r^2 - 2r(k^2-\cos\theta)+(k^2-1)=0$, has a coefficient of $r$ equal to $-2(k^2-\cos\theta)$ — a constant term $-2k^2$ plus a $\cos\theta$ term, which a true circle's equation cannot have. Concretely, if the region were the circle through the two real-axis endpoints $x_{\min}=-\tfrac13$ and $x=1$ (for $k=2$), it would need center $\tfrac13$ and radius $\tfrac23\approx0.667$, giving $y_{\max}=\sqrt{0.667^2-0.333^2}\approx0.577$ — but the true value is $y_{\max}\approx0.451$. So the region is a closed, egg-shaped oval (flatter than the naive circle guess), not a genuine circle. (For reference, the boxed "Apollonius circle" formula sometimes quoted for this problem, $|z-\tfrac{k^2}{k^2-1}|<\tfrac{k}{k^2-1}$, is centered outside the unit disk for every $k>1$ and does not even contain $z=0$ — which we showed above always satisfies the inequality — so that formula is incorrect.)