I want to show that that $$\lim_{z\to\infty}|\sum_{k=1}^n p_k(z)e^{\lambda_k z}|=0$$ iff $\Re(\lambda_k)<0$ for all $k=1,\dots,n$ where $p_k\in\mathbb C[z]\backslash \{0\}$ and $\lambda_k\in\mathbb C$ with pairwise different real parts.
My thoughts so far:
At first I considered the individual summands: For $z=x+iy$ we have $$|p(z)e^{\lambda z}|=|p(z)||e^{\Re(\lambda)\cdot x-\Im(\lambda)\cdot y}|\cdot\underbrace{|e^{i(\Re(\lambda)\cdot y+\Im(\lambda)\cdot x)}|}_{=1},$$so $$|p(z)e^{\lambda z}|=|p(z)|\frac{e^{\Re(\lambda)\cdot x}}{e^{\Im(\lambda)\cdot y}}.$$ Now, is there a way I can finish this argument to obtain that $\lim_{z\to 0}|p(z)e^{\lambda z}|=0$ iff $\Re(\lambda)<0$? And if so, can I then say that $$|\sum_{k=1}^n p_k(z)e^{\lambda_k z}|\leq\sum_{k=1}^n|p(z)e^{\lambda_k z}|\to 0$$ iff $\lambda_k <0$ for all $k=1,\dots,n$? What confuses me is that for some reason it is necessary that the $\lambda_k$ must have pairwise different real parts.
This is not true. Take $n=1 , p_1 (z) =1 , \lambda_1 = -1 .$ Let $z_m = im , m\in \mathbb{N} .$ Then $z_m \to \infty$ but $$\left|\sum_{k=1}^1 p_1 (z_m ) e^{\lambda_k z_k} \right| = |e^{-ik} | =1$$