Real roots of partial sums of analytic function represented by infinite series : Consider the following polynomial:
$$P_n(x)= \sum_{k=1}^n a_kx^k$$
We know (and can verify) that , for every $n$ , $P_n(x)$ has only real roots 
Now, define
$$P(x)= \sum_{k=1}^\infty a_kx^k$$
We know that $P(x)$ analytic and entire .


Question : 
Does this mean that $P(x)$ has only real roots ? If not please elaborate with examples 


Note: I'm not sure this is MSE or MO question (?)
Edit: Converse of the above hypothesis isn't true
 A: Let us assume that $P(z)$ has an root $z_0$, $\Im(z_0) \neq 0$ (without a loss of generality, we can assume $\Im(z_0)>0$). Since $P(z)$ is an entire function, this root is necessarily isolated (unless $P(z) \equiv 0$, but then $P_n(z) \equiv 0$ doesn't fullfill the assumptions). Let then $\gamma$ be a closed curve surrounding $z_0$, not surrounding any other of roots of $P(z)$, and entirely enclosed within the semiplane $\Im(z)>0$.
Then, for any $k\in\mathbb N$, $\frac{(z-z_0)^k}{P_n(z)}$ is holomorphic in the region bounded by $\gamma$, while $\frac{(z-z_0)^k}{P(z)}$ has at most a single singularity within, at $z_0$. We have
$$0 = \int_\gamma \frac{(z-z_0)^k}{P_n(z)} dz = \lim_{n\rightarrow\infty} \int_\gamma \frac{(z-z_0)^k}{P_n(z)} dz = \int_\gamma \frac{(z-z_0)^k}{P(z)} dz = 2\pi i {\rm Res}_{z_0} \frac{(z-z_0)^k}{P(z)}$$
Since this is true for any $k\in\mathbb N$, that means that the whole part of Laurent series of $\frac{1}{P(z)}$ around $z_0$ with negative powers of $(z-z_0)$ vanishes, i.e. $\frac{1}{P(z)}$ doesn't have a singularity at $z_0$. This contradicts the assumption that $P(z)$ has a root at $z_0$.
Therefore, $P(z)$ cannot have non-real roots.
