$1-2^z+3^z-4^z=0$ with $z\in\mathbb{C}$ I'm looking for characteristics of the solutions for
$$1-2^z+3^z-4^z=0$$ with $z\in\mathbb{C}$. 
I want to compare it with the simple case $1-2^z=0$. The problem is, that I haven‘t the technical possibility to check this, therefore this post.
Can somebody help me to find out, if
(1)   there are infinite complex zeroes (I think so) ,
(2)   the real part of all complex zeroes ($\neq 0$) is the same (here: $\Re(z)\neq 0$ is constant and maybe negative) .
[Addition: Also the solutions of $1+2^z+3^z+4^z=0$ are interesting.]
Thanks in advance.     
 A: Yes there are infinitely many zeroes, and no their real parts aren't all the same.
If $r_- < 0$ is the solution to $1 = 2^r+3^r+4^r$ and $r_+ > 0$ is the solution to $4^r = 1+2^r+3^r$, then you have $|1-2^z+3^z-4^z| \ge 1 - (|2^z|+|3^z|+|4^z|) > 0$ for $\Re(z) < r_-$ and $|1-2^z+3^z-4^z| \ge |4^z| - (1+|2^z|+|3^z|) > 0$ for $\Re(z) > r_+$, and so the solutions are located in the vertical band $r_- \le \Re(z) \le r_+$.
Using the residue theorem from complex analysis, the number of solutions with $|\Im(z)| \le Y$ is $\frac 1 {2i\pi}\int_{\gamma(Y)} \frac {f'(z)}{f(z)} dz$ where $\gamma(Y)$ is any rectangular contour whose top and bottom edges are at $\Im(z) = \pm Y$, and the left and right edges are beyond $r_-$ and $r_+$.
The integral on the left edge gets as close to $0$ as you want when you move it enough to the left, the integral on the right edge gets to $2iY\log(4)$ which gives a contribution of $Y\log(4)/\pi$.
Meanwhile, the integral on the horizontal edges give a mostly real contribution when $\Re(z)$ is away from $[r_-;r_+]$, which doesn't add anything to the count.
If you expect that $\int_{r_-+iY}^{r_++iY} \frac {f'(z)}{f(z)}dz$ is bounded as $Y$ varies, you should have that the number of zeros in the band with imaginary part bounded by $Y$ is $Y \log(4)/\pi + O(1)$.
In the case where you have one less exponential term, one can show that the sequence of the real parts behave like a pseudo-random sequence whose distribution have an explicit closed-form, but here I'm really not sure if that's still the case. Since $\log 4 / \log 2$ is rational, in fact there is still a chance that there is something.

Here is an argument showing that $r_-$ is the best limit :
Since $\log(2)/\log(3)$ is irrational it's possible to find a $y$ such that $y\log(2) \approx 0 \pmod {2\pi}$ and $y\log(3) \approx \pi \pmod {2\pi}$ : just try all the $y = 2k\pi/\log(2)$. Then $\log(3)y = 2k\pi\log3/\log(2)$. Since the sequence $(k\log(3)/\log(2) \pmod 1$ is dense in the unit interval, it gets arbitrarily close to $1/2$. So there is a sequence $(y_n)$ such that $y_n\log(2) = 0 \pmod {2\pi}$ and $\varepsilon_n = (y_n\log(3) \pmod {2\pi}) - \pi \to 0$.
Now take $z$ near $z_n = r_- + iy_n$, i.e. $z = (r_-+iy_n)+dz$.
Doing a development at order $1$ and using $1-2^{r_-}-3^{r_-}-4^{r_-} = 0$, we have $f(z) = 1 - 2^z + 3^z - 4^z = (- \log 2. 2^{r_-} - \log 3. 3^{r_-} - \log 4. 4^{r_-}) dz - \log 3. 3^{r_-}(i\varepsilon_n) + O((dz,\varepsilon_n)^2)$
Now if you have $z$ move in a tiny circle around $z_n$ and if $\varepsilon_n$ is small enough, the error term on that circle is too small compared to this approximation of $f(z)$, which shows that they have the same number of zeroes (exactly one) .
The $r_+$ bound is also the best, this time you want to have $2^z$ and $3^r$ almost a negative real simultaneously. Then $4^z$ is almost a positive real and you are in the situation where $1,-2^z,-3^z$ are aligned against $-4^z$
