sequences and series - convergence criteria from real to complex domains?

It is well known the following function

$$f(x) = \frac{1}{1^x} + \frac{1}{2^x} + \frac{1}{3^x} + \frac{1}{4^x} + \cdots$$

only converges if $x > 1$.

If we now consider $f(z)$ where z is complex, why can we say that the function converges for $Re(z) > 1$?

What's logic that allows us to simply apply convergence rules to the real part of a complex function's domain?

(I am not a trained mathematician so I'd appreciate answers which minimise assumptions about terminology.)

Answer

As indicated by Martin R in the comment, the reason is that absolute convergence of complex series implies convergence and in this case we have that for $z=x+iy$

$$\left|\frac1{n^z}\right|=\frac1{|n^z| }=\frac1{|n^{x}|}$$

indeed

$$|n^z|= |n^x||n^{iy}|=|n^x||e^{iy\log n}|=|n^x|$$

thus the series conveges for $Re(z)=x>1$.