For a given sequence of complex functions: $\phi_n(z)= 1+\frac1n-z-e^{-z}$; here $z\in${$z| Rez>0$}.
I want to prove that :
(1). $\phi_n $ has a unique zero $z_n$ in the half plane. (i.e. there exists a unique $z_n$ in half plane {$z| Rez>0$} s.t $\phi_n(z_n)=0)$.
[Some how by Rouche's theorem, I almost proved that the function $1+\frac1n-z-e^{-z}$ has only one zero in half plane {$z| Rez>0$}. But I don't know how to show that $z_n$ is unique s.t. $f_n(z_n)=0$ ]
(2). And further more, I have to show that $z_n\in \Bbb R$.
(3). Does $lim_{x\to \infty} z_n$ exist? And If limit exists then what is it?
(I don't have any idea about last two parts.)
Thanks in advance.
I would suggest that you start by showing that each $\phi_n$ has a positive real zero. That is an easy consequence of the intermediate value theorem, since all $\phi_n$ are real-valued on $\mathbb{R}$. That way, you are not distracted by thinking about the uniqueness.
Then show that each $\phi_n$ has only one zero in the right half plane. Rouché's theorem is a good option for that, one just has to find the right bounded subregion for it to be applicable. For that, note that no $\phi_n$ can have a zero $z_0$ with $\operatorname{Re} z_0 \geqslant 3$ (arbitrarily chosen value, others work too, just not too small) or $\operatorname{Im} z_0 \geqslant 3$ by the triangle inequality.
For the application of Rouché's theorem, let us choose the rectangle $R$ with vertices $-i\pi,\, \pi - i\pi,\, \pi +i\pi,\, i\pi$. All zeros of $\phi_n$ in the right half plane must lie in the interior of $R$, because
$$\lvert \phi_n(z)\rvert = \left\lvert 1 + \frac1n - z - e^{-z}\right\rvert \geqslant \lvert z\rvert - 1 - \frac1n - \lvert e^{-z}\rvert \geqslant \lvert z\rvert - 3 \geqslant \pi-3 > 0$$
for $\lvert z\rvert \geqslant \pi$ and $\operatorname{Re} z \geqslant 0$.
We apply Rouché's theorem to $f_n(z) = \frac1n - z$ and $g(z) = 1 - e^{-z}$. On the three sides of $\partial R$ that lie in the right half plane, we have
$$\lvert g(z)\rvert \leqslant 1 + \lvert e^{-z}\rvert \leqslant 2 < \pi - 1 \leqslant \lvert z\rvert - \frac1n \leqslant \lvert f_n(z)\rvert.$$
On the side of the rectangle that lies on the imaginary axis, we compute in a different way, we have
$$\left\lvert 1 - e^{-it}\right\rvert^2 = \left\lvert (1-\cos t) + i\sin t \right\rvert^2 = (1-\cos t)^2 + \sin^2 t = 2(1-\cos t) = 4 \sin^2 (t/2),$$
and
$$\left\lvert\frac1n - it \right\rvert^2 = \frac{1}{n^2} + t^2 > t^2 = 4(t/2)^2 \geqslant 4\sin^2(t/2).$$
So we have established that $\lvert g(z)\rvert < \lvert f_n(z)\rvert$ on $\partial R$, and can apply Rouché's theorem to the functions $f_n$ and $\phi_n = f_n+g$ on the rectangle, and conclude that both have the same number of zeros inside $R$. Evidently, $f_n$ has the single zero $\frac1n$ in $\mathbb{C}$, which lies inside $R$, so $\phi_n$ also has a unique zero in $R$.
Finally, for the third point, knowing that the zeros $z_n$ are real allows to approximately locate them with ease, finding $a_n,b_n > 0$ such that $a_n < z_n < b_n$ and such that $\lim (b_n-a_n) = 0$. Then you can see whether $\lim a_n$ exists. $\lim z_n$ exists if and only if $\lim a_n$ exists, and then the two are equal.