Non-real roots of $z^2=\sin(z)$ 
What is the number of roots of $$z^2= \sin(z)$$ in $\left\{z\in\mathbb C\setminus\mathbb R\left||z|<2\right.\right\}$?

This task is related to the Argument principle and Rouché's theorem.
Maybe I should estimate $f(z)=z^2$ and $g(z)= \sin(z)$ using an inequality that would hold when $|z|<2$.
I would be grateful if you provide an explanation, so I could learn how to solve similar problems.
 A: Consider $g(z)=\sin z-z^2$ and $D=\{z\in\mathbb{C}:\lvert z\rvert<2\}$. I claim that 
$$\bbox[10px,#ffd]{
\lvert \sin z\rvert<\lvert z\rvert^2
}
$$
on $\partial D$. From this approach, if we let $z=x+iy$, then 
$$
\max_{\lvert z\rvert=2}\, \lvert\sin z\rvert=\max_{x^2+y^2=4}\,\left\lvert\sqrt{\cosh^2y-\cos^2x}\right\rvert
$$
The last function has (on the constrained domain) two maxima at $(0,2)$ and $(0,-2)$, which is equivalent to $\pm2i$. Evaluating at these values yields 
$$
\lvert\sin(\pm2i)\rvert=\sqrt{\cosh(2)^2-1}\approx3.62
$$
So on $\partial D$, 
$$
\lvert\sin z\rvert\leq 3.63<4=\lvert z\rvert^2
$$
and our claim is justified. Now, by Rouche's theorem, $z^2$ and $g(z)$ have the same number of zeros (counted with multiplicity) in $D$ - namely 2. 
Now, we consider the equation
$$
\sin z=z^2
$$
for $z\in\mathbb R$ only. Since $|\sin(z)|\le 1$ for $z\in\mathbb R$, we must have $z\in[-1,1]$. Now we note that $z=0$ is a real solution. Since $$(0.1)^2<\sin(0.1) \text{ and } 1^2=1>\sin(1),$$ we know from the Intermediate value Theorem that there is another solution for some $0.1<z<1$. 
Hence, both solutions in $D$ are real and it follows that there are no non-real solutions to $\sin(z)=z^2$ on $D$.
