Question: > 
> Is it always valid that
>  $$ z^{a+ib} = z^a\cdot z^{ib} \qquad \forall z\in \mathbb{C}.$$
> 
> 
> 

In high school I was always taught to see the $+$ in complex numbers as analogous to that is reals. However can it be proven to be valid always, if at all it can be ? Also what is the intuition behind complex exponentiation? (It does not seem to be as straightforward as with reals)

Answer: 
We define complex power functions using the exponential and logarithm functions:
$$
z^\alpha = e^{\alpha\log z}.
$$
The problem with this definition is that the complex logarithm function cannot be defined continuously on the entire punctured complex plane. (The punctured plane is the plane with zero removed. Of course you cannot define the logarithm to be continuous at zero, but there is a deeper problem. By contrast, the exponential function *is* defined continuously on the entire complex plane, so that part of the formula doesn't give a problem.) To explain why, switching to polar coordinates and writing $z=r e^{i\theta}$ gives
$$
\log z = \log r + i\theta.
$$
But in general, there is no continuous way to assign a [complex argument](https://en.wikipedia.org/wiki/Argument_(complex_analysis)) (the angle $\theta$ above) to every complex number in the punctured plane.

More specifically, it is clear from the formula above that all possible values of $\log z$ differ by an integer multiple of $2\pi i$. So, we can say informally that the logarithm function is multi-valued:
$$
\log z = \log r + i\theta + 2ni\pi.
$$
Moving back to the definition above of power function, this means that the power function $z^\alpha$ is multi-valued:
$$
z^\alpha = e^{\alpha\log z} = e^{2n\pi i\alpha}e^{\alpha\log z} = e^{2n\pi i\alpha}z^\alpha.
$$
So in other words, $z^\alpha$ is only defined up to a factor of powers of $e^{2\pi i\alpha}$. For example, the possible values of $i^i$ are
$$
\dots,e^{-3\pi},e^{-\pi},e^\pi,e^{3\pi},\dots.
$$
If $\alpha$ is an integer then $2n\pi i\alpha$ is an integer multiple of $2\pi i$ (since $e^{2\pi i}=1$; this is related to [Euler's formula](https://en.wikipedia.org/wiki/Euler%27s_formula)) and so $z^\alpha$ has only one possible value; this is why we can define e.g. $z^{45}$ without any ambiguity.

So to finally answer your question, the formula $z^{a+bi} = z^a\cdot z^{bi}$ *is* valid as long as we choose the right values for $z^{a+bi}$, $z^a$, and $z^{bi}$, thinking of these as multi-valued functions.
The law of exponents
$$
x^ax^b=x^{a+b}
$$
Remains valid. However, as you note, exponentiation is a bit more complicated and is defined as follows, for a complex number $s$ and a real base $a$:
$$
a^s=e^{s\log n}
$$
As is seen in the (super important) class of series called Dirichlet series, for $a$ natural, a special case of this being the Riemann Zeta function.

The above gives you the tools to tackle expressions such as
$$
z^s,\Im(s),\Im(z)\ne0
$$
meaning both $z$ and $s$ have imaginary components. Representing $z$ in polar form
(as you can always do in the complex plane) gives
$$
z=re^{i\theta}\Rightarrow z^{s}=r^{s}e^{si\theta}
$$
Since $r$ is real, we can tackle it given our previous definition. If we take $s=a+bi$, then the exponential is just
$$
e^{si\theta}=e^{(a+bi)i\theta}=e^{ai\theta-b\theta=}e^{ai \theta}e^{-b\theta}
$$
Where in the last step, we broke it up again so we can absorb $e^{-b\theta}$ into the new radius of $z^s$. 

Hope this helps!