Is there $z \in \mathbb{C}$ with $\mathrm{Re}(z) > 0$ such that $e^{2z}+\frac{z+1}{z-1} = 0$? Let $z$ be a complex number with positive real part. Is it possible that $e^{2z}+\frac{z+1}{z-1} = 0$?
Define:
$$f:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto(z-1)e^{2z}+(z+1),$$
$$g_1:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto|f(x+iy)|^2,$$
$$g_2:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto g(x,y)e^{-2x},$$
$$h:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto(x-1)e^{2x}+1.$$
Now prove this:
$$(\forall x&gt;0)\frac{\partial g_1}{\partial x}&gt;0\Rightarrow(\forall x&gt;0)g_1(x,y)&gt;0,$$
$$(\forall x&gt;0)\frac{\partial g_2}{\partial x}&gt;0\Rightarrow(\forall x&gt;0)\frac{\partial g_1}{\partial x}&gt;0,$$
$$(\forall x&gt;0)\frac{\partial g_2}{\partial x}\geqslant 2x(h(x)+h(-x)),$$
$$(\forall x\neq 0)h(x)+h(-x)&gt;0.$$
Near $z=1$, the second term dominates the first; far from $z=1$, the first dominates the second.  By numerical methods, the curve in the right half plane where $\left|e^{2z}\right| = \left|(z+1)/(z-1)\right|$ looks like this:

and nowhere on that curve (except the origin) is there a zero of your function.