The second coefficient of a normalized univalent function has modulus at most two

Since $f$ is injective and $f(0)=0$, the only zero of $f$ in $\mathbb{D}$ is $0$. Hence $F(z):=\frac{f(z^2)}{z^2}=1+a_2{z}^2+a_3{z}^4+\cdots$ extends holomorphically and nowhere vanishingly to $\mathbb{D}$. By [L2] and [L3], choose a holomorphic square root $q$ on $\mathbb{D}$ with $q(z{)}^2=F(z)$ and $q(0)=1$.

Put $h(z):=zq(z)$. Then $h(z{)}^2=f(z^2)$. The function $h$ is odd and univalent: if $h(z_1)=h(z_2)$ then $f(z_1^2)=f(z_2^2)$, so $z_1^2=z_2^2$; if $z_1=-z_2\mathrm{\ne }0$, oddness gives $h(z_1)=-h(z_2)$, contradiction. Thus $h(z)=z+\frac{a_2}{2}{z}^3+\cdots \mathrm{.}$

Define $G(\zeta ):=\frac{1}{h(\zeta )}=\frac{1}{\zeta }-\frac{a_2}{2}\zeta +\cdots (0<\mathrm{\mid }\zeta \mathrm{\mid }<1)\mathrm{.}$ Since $h$ is injective on $\mathbb{D}$ and vanishes only at $0$, the map $G$ is holomorphic and injective on the punctured disc. Fact [L1] therefore applies and yields ${\mid \frac{a_2}{2}\mid }^2\le 1.$

Therefore $\mathrm{\mid }{a}_2\mathrm{\mid }\le 2$.