Consider the two closely related polynomials,
$$P_1(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_nx^n$$
$$P_2(x)=a_0-a_1x+a_2x^2-a_3x^3\cdots+(-1)^na_nx^n$$
I have noticed, by plugging in numbers, that the roots of $P_1$, $\{z_0,z_1,\cdots ,z_n\}$, are related to the roots of $P_2$, $\{\tilde{z}_0,\tilde{z}_1,\cdots ,\tilde{z}_n\}$, by the following relation:
$$\Re(z_i)=-\Re(\tilde{z}_i)$$
$$\Im(z_i)=\Im(\tilde{z}_i)$$
This seems so simple to prove, but I just can't prove it. Could you help me out.
On
First of all, I would suggest a change of notation in your question as $\bar{z}$ represents the conjugate of $z$.
In fact, if $z$ is a solution of $P_1(x)=0$ (with all real coefficients), then so is $\bar{z}$.
Now note that if $x$ solves the first equation, then $-x$ solves the second.
This is because $(-1)^{2k}=1$ whereas $(-1)^{2k+1}=-1$
Also note that if $z = -w$, where $z$ and $w$ are both complex numbers, $\mathrm{Re}(z)=-\mathrm{Re}(w)$ and $\mathrm{Im}(z)=-\mathrm{Im}(w)$ because $-(a+bi) = -a-bi$.
If we let $z$ and $w = -z$ be respective solutions of $P_1(x)$ and $P_2(x)$, then so are $\bar{z}$ and $\bar{w}$ (respectively). Hence what you're observing (and comparing) is the pair $z$ and $\bar{w}$. But those are not the only solutions.
Notice that $P_2(x)=P_1(-x)$ so the roots of $P_2$ are just the negatives of the roots of $P_1$. If the coefficients of $P_1$ are real, the roots of $P_1$ can also be described as the complex conjugates of the roots of $P_1$. Thus in that case we can describe the roots of $P_2$ as the negatives of the complex conjugates of the roots of $P_1$, which is exactly the relationship you wrote.