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  "body": "\n## Proof\n\n1. Let $f$ and $g$ be [[holomorphic|0DG0Z72CQD]] on an open set $W$ that contains $\\set{z : \\abs{z - c} \\le r}$. Let $C$ be the circle $\\abs{z - c} = r$, and assume $\\abs{g} \u003c \\abs{f}$ on $C$.\n2. For $t \\in [0, 1]$, let $F_t = f + t g$. It is holomorphic on $W$, so it is [[meromorphic|0NSD1TPHR0]] on $W$ with no poles.\n3. On $C$, the [[reverse triangle inequality|0C9S13WKMF]] gives $\\abs{F_t} \\ge \\abs{f} - t \\abs{g} \\ge \\abs{f} - \\abs{g}$.\n4. The circle $C$ is compact by the [[Heine–Borel theorem|07PFYV65MB]]. By [[attainment of bounds|0V4DET59NW]], $\\abs{f} - \\abs{g}$ has a minimum $m$ on $C$, and $m \u003e 0$ by hypothesis. So $\\abs{F_t} \\ge m$ on $C$.\n5. Let $Z(t)$ be the number of zeros of $F_t$ in $\\abs{z - c} \u003c r$, counted with [[order|064N002TX4]]. By step 4 and the [[argument principle|0PN2BXRS8G]], $Z(t) = \\frac{1}{2\\pi i} \\oint_C \\frac{F_t'}{F_t} \\, dz$, a [[circle integral|0T7ABX4CWE]]. In particular $Z(t)$ is an integer.\n6. For $s, t \\in [0, 1]$, a direct expansion gives $(f' + t g')(f + s g) - (f' + s g')(f + t g) = (t - s)(g' f - f' g)$.\n7. Dividing by $F_t F_s$ gives $\\frac{F_t'}{F_t} - \\frac{F_s'}{F_s} = \\frac{(t - s)(g' f - f' g)}{F_t F_s}$ on $C$.\n8. By [[holomorphic functions are infinitely differentiable|0QEBN8DGVE]], $f'$ and $g'$ are holomorphic, hence continuous. So $\\abs{g' f - f' g}$ has a maximum $K_0$ on $C$.\n9. By steps 4 and 7, $\\abs{F_t'/F_t - F_s'/F_s} \\le \\abs{t - s} K_0 / m^2$ on $C$. The length of $C$ is $2\\pi r$.\n10. By the [[estimation lemma|0DJMGPN8K9]] and step 5, $\\abs{Z(t) - Z(s)} \\le \\frac{1}{2\\pi} \\cdot 2\\pi r \\cdot \\frac{\\abs{t - s} K_0}{m^2} = \\frac{r K_0}{m^2} \\abs{t - s}$.\n11. Choose an integer $q \\ge 1$ with $q \u003e r K_0 / m^2$, and let $t_j = j/q$ for $j = 0, \\dots, q$. By step 10, $\\abs{Z(t_{j+1}) - Z(t_j)} \u003c 1$.\n12. Since the values of $Z$ are integers, $Z(t_{j+1}) = Z(t_j)$ for each $j$. Hence $Z(0) = Z(1)$.\n13. Here $F_0 = f$ and $F_1 = f + g$. So $f$ and $f + g$ have the same number of zeros in $\\abs{z - c} \u003c r$, counted with order.\n\n## Idea\n\nMove continuously from $f$ to $f + g$ through $f + t g$. The condition $\\abs{g} \u003c \\abs{f}$ on the circle keeps every $f + t g$ away from zero there. So the argument principle counts the zeros of each $f + t g$ by an integral that depends continuously on $t$. An integer that varies continuously cannot change.\n",
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