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Let $D=\{z: |z|<1\}$ and suppose that $f:D\rightarrow \mathbb{C}$ is analytic, one-to-one and $f(0)=0, f'(0)=1.$ Show that there exists $g:D\rightarrow \mathbb{C}$ is analytic and one-to-one such that $(g(z))^2=f(z^2).$

I proved this statement is as follows:

Since $f$ is analytic and $f(0)=0, f'(0)=1$, we have $f(z)=z+a_1z^2+a_2z^3+...$

Then, $f(z^2)=z^2(1+a_1z^2+a_2z^3+...)=z^2h(z).$ If there is $z_0\in D$ such that $h(z_0)=0$, we have $f({z_0}^2)=0$, which implies $z_0^2=0$ since $f$ is 1-1. Thus, $z_0=0.$ However, it is clear that $h(0)=1\neq 0.$ Hence, we proved that $h$ is nonzero on $D$.
Now, since $h$ and $\frac{1}{h}$ are both analytic on $D$, there is an analytic function $g_1:D\rightarrow \mathbb{C}$ such that $g_1^2=h.$ (this result follows from the theorem 13.11 in Real and Complex analysis by W. Rudin) Therefore, we define $g(z)=zg_1(z)$, it is clear that $g$ is analytic and satisfies $g(z)^2=f(z^2).$

However, I have no good idea how to explain that $g$ is 1-1. Can anyone give me some ideas how to solve this part? Thanks.

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    Your proof is good. you used a theorem mentioned in rudin, also say $$g(z)=\sqrt{f(z^2)}=\sqrt{z^2(1+a_1z^2+a_2z^3+\cdots)}=z+\dfrac12a_1z^3+\cdots$$ – 2017-02-09
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    Thank you to your answer. I wonder how can we get the last equality? – 2017-02-09

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Note that $$ g(z)=\sqrt{f(z^2)}=z\sqrt{1+a_1z^2+a_2z^3+\cdots}=z+\dfrac12a_1z^3+\cdots $$ is an odd function and if $g(z_1)=g(z_2)$ for $z_1,z_2\in{\mathbb D}=\{z: |z|<1\}$ then $f(z_1^2)=f(z_2^2)$ or $z_1=\pm z_2$, but if $z_1=-z_2$ then $$g(z_1)=g(-z_2)=-g(z_2)=-g(z_1)$$ where $g\neq0$ thus $g$ is one-to-one.

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To MyGlasses. Thank you a lot. That is very helpful. I found another way to illustrate $g$ is odd. I tried to prove this result by showing that $g_1$ is even. Since $h(z)=\frac{f(z^2)}{z^2}$, and therefore $h$ is even. On the other hand, $g_1(z)=e^{1/2 Log h(z)}$. It is clear that $g_1(z)=g_1(-z)$, so $g_1$ is also even. Hence $g(z)=zg_1(z)$ is odd.