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  "body": "\n## Proof\n\n1. Let $\\mathbb{D}$ and $f$ be as in the statement. By [[holomorphic functions are analytic|0826SGV3R0]] with $c = 0$ and radius $1$, $f(z) = \\sum_{n \\ge 0} a_n z^n$ for each $z \\in \\mathbb{D}$.\n2. We have $a_0 = f(0) = 0$. By [[power series are holomorphic|0YMP521ZR5]], differentiating term by term at $0$ gives $f'(0) = a_1$.\n3. The series $\\sum_{n \\ge 1} a_n z^{n-1}$ converges at each $z \\in \\mathbb{D}$: for $z \\ne 0$ its partial sums are $1/z$ times those of the series for $f(z)$.\n4. A power series that converges at $w$ has bounded terms there. So, by the definition of the [[radius of convergence|045EWAXY9P]], the series of step 3 has radius at least $1$.\n5. Let $g(z) = \\sum_{n \\ge 1} a_n z^{n-1}$ on $\\mathbb{D}$. By [[power series are holomorphic|0YMP521ZR5]], $g$ is [[holomorphic|0DG0Z72CQD]] on $\\mathbb{D}$, hence continuous.\n6. By steps 2 and 3, $g(z) = f(z)/z$ for $z \\ne 0$, and $g(0) = a_1 = f'(0)$.\n7. Claim: for $0 \u003c r \u003c 1$ and $\\abs{z} \\le r$, $\\abs{g(z)} \\le 1/r$. The disc $K_r = \\set{z : \\abs{z} \\le r}$ is compact by the [[Heine–Borel theorem|07PFYV65MB]].\n8. By [[attainment of bounds|0V4DET59NW]], $\\abs{g}$ has a maximum on $K_r$ at some $z^*$.\n9. If $\\abs{z^*} \u003c r$, then $\\abs{g}$ has a local maximum at $z^*$ in the domain $\\mathbb{D}$. By the [[maximum modulus principle|048F2P90WR]], $g$ is constant, so the maximum is also attained on $\\abs{z} = r$.\n10. So in all cases the maximum of $\\abs{g}$ on $K_r$ is attained at a point $z'$ with $\\abs{z'} = r$. There $\\abs{g(z')} = \\abs{f(z')}/r \u003c 1/r$, since $f$ maps into $\\mathbb{D}$. This proves the claim.\n11. Fix $z \\in \\mathbb{D}$. Step 7 holds for every $r$ with $\\abs{z} \\le r \u003c 1$. Letting $r \\to 1$ gives $\\abs{g(z)} \\le 1$.\n12. For $z \\ne 0$, step 11 gives $\\abs{f(z)} \\le \\abs{z}$, and this also holds at $z = 0$. At $z = 0$ it gives $\\abs{f'(0)} \\le 1$.\n13. Now assume $\\abs{f(z_0)} = \\abs{z_0}$ for some $z_0 \\ne 0$, or $\\abs{f'(0)} = 1$. Then $\\abs{g(z_0)} = 1$ or $\\abs{g(0)} = 1$.\n14. By step 11, $\\abs{g} \\le 1$ on $\\mathbb{D}$. So $\\abs{g}$ has a local maximum at an interior point of $\\mathbb{D}$.\n15. By the maximum modulus principle, $g$ is a constant $a$. By step 13, $\\abs{a} = 1$.\n16. Then $f(z) = z g(z) = a z$ for $z \\ne 0$, and also for $z = 0$ since $f(0) = 0$.\n\n## Idea\n\nDividing by $z$ removes the zero at the origin and gives a holomorphic $g$ with $g(0) = f'(0)$. On the circle $\\abs{z} = r$ we have $\\abs{g} \u003c 1/r$, and the maximum modulus principle carries this bound inside. Letting $r \\to 1$ gives $\\abs{g} \\le 1$, and equality at one interior point forces $g$ to be a constant of modulus one.\n",
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