= Solution
Yes. An explicit <holomorphic> surjection is the <rational function>
$$
\boxed{F(z)=\left(\frac z{1-z}\right)^2,\qquad |z|<1}.
$$
Its denominator never vanishes inside the <unit disc>. To prove surjectivity, take any $w\in\mathbb C$ and choose a square root $r$ of $w$ with $\operatorname{Re}r\geq0$; one of the two opposite roots always has this property. Define $z=r/(1+r)$. Then $1+r\ne0$ and
$$
|1+r|^2-|r|^2=1+2\operatorname{Re}r>0,
$$
so $|z|<1$. Also $z/(1-z)=r$, and consequently $F(z)=r^2=w$.
This is a <holomorphic surjection from the unit disc onto the complex plane>. It does not contradict part (i): it is not injective. For example, $F(i/(1+i))=F(-i/(1-i))=-1$, with two distinct arguments of modulus $1/\sqrt2$.
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