Consider the $T : z \to z+a$ for $a \in \mathbb{C}$ (Here the $a = \alpha + i \beta$)
Take any complex function, $f(z)$
Then, $T \circ f(z) = f(z) + a$ is translation by $a$ (I.e. $f(z)$ is translated by $\alpha$ to x-axis and $\beta$ to y-axis)
But What is the exact meaning of the $f \circ T = f(z+a)$?
To find the answer, I took the $f(x)$ and $f(x,y)$ for real variable $x$ and $y$ with the $T_1 : x \to x+\alpha$ and $T_2 : (x,y ) \to (x+\alpha, y+\beta)$
Then, In my thought
$f \circ T_1 (x) = f(x-\alpha)$ since the $f$ is translated by "$\alpha$" for x-axis
Like the above, $f \circ T_2(x,y) = f(x-\alpha,y-\beta)$ with the same reason for case $f(x-\alpha)$.
So the $f \circ T(z) = f(z+a)$ would be translated by "$-a$" for $f(z)$ if it was regarded like the real case. But the $T$ itself, translated by "$a$" not the "$-a$". It is contradiction.
Hence I'm very confused between real and complex translation for $f(z+a)$. What is the difference between real and complex for that? Is the $f(z+a)$ transition by "$a$", "$-a$" or neither? Is it similar with the real case?
Write $f(z)=u(x,y)+iv(x,y)$, where $z=x+iy$. Assume $a=\alpha + i\beta$. Then $f(z+a)=f((x+\alpha)+i(y+\beta))=u(x+\alpha,y+\beta)+iv(x+\alpha,y+\beta)$.
In real variables, "+" general means shifted left for the horizontal variable ($x$) and down for the vertical variable ($y$). We can understand negative numbers as negative shifts by this convention. For example, for some real function $g(x)$, we can say $g(x-5)$ is $g$ shifted left by $-5$, which is the same as shifting right by $5$.
Anyways, from the equation I wrote, it is clear that the real and imaginary parts are being shifted "left" by $\alpha$ and "down" by $\beta$, where a negative shift is possible.