Why do logarithms and arguments (complex numbers) have similar properties? Why does the logarithm and argument (complex numbers) have similar properties such as $$\log(xa)=\log(x)+\log(a), \arg(xa)=\arg(x)+\arg(a)\\ \log(\frac{x}{a})=\log(x)-\log(a), \arg(\frac{x}{a})=\arg(x)-\arg(a)$$
I understand the proof of such items individually, but on more of an intuitive level why is this the case?
What is this property of the functions called, where it almost "reduces" the operation such as powers to multiplying, multiplying to adding etc.?
Thanks
 A: The "intuitive" reason you seek, is that any complex number $z$ can be written in function of its argument $\theta$ as follows:
$$z = re^{i\theta},$$
where $r$ is the modulus of $z$.
Now, multiplying two complex numbers $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$ gives us
$$z_1z_2 = r_1r_2e^{i(\theta_1+\theta_2)}$$
The reason the argument behaves like a logarithm, is because it is situated in the exponent of $e$, or conversely, because 
$$\log(z) = \log(r) + i\theta,$$
meaning that the argument $\theta$ is the imaginary part of $\log(z)$.
NOTE: this logarithm is the complex logarithm and is a multi-valued function. Everything to do with the argument only holds up to multiples of $2\pi$, so be careful with claims about equality!
A: When $x>0$:
$$\ln(x)\equiv\int_1^x\frac{1}{t}\space\text{d}t$$
Now, when $x=\text{a}\cdot\text{b}$:
$$\ln(\text{a}\cdot\text{b})=\int_1^{\text{a}\cdot\text{b}}\frac{1}{t}\space\text{d}t=\int_1^{\text{a}}\frac{1}{t}\space\text{d}t+\int_1^{\text{b}}\frac{1}{t}\space\text{d}t$$
And, when $\text{z}_1\space\wedge\space\text{z}_2\in\mathbb{C}$:
$$\arg\left(\text{z}_1\cdot\text{z}_2\right)=\arg\left(\left|\text{z}_1\right|e^{\arg\left(\text{z}_1\right)i}\cdot\left|\text{z}_2\right|e^{\arg\left(\text{z}_2\right)i}\right)=\arg\left(\left|\text{z}_1\right|\cdot\left|\text{z}_2\right|\cdot e^{\arg\left(\text{z}_1\right)i+\arg\left(\text{z}_2\right)i}\right)=$$
$$\arg\left(\left|\text{z}_1\right|\cdot\left|\text{z}_2\right|\cdot e^{\left(\arg\left(\text{z}_1\right)+\arg\left(\text{z}_2\right)\right)i}\right)=\arg\left(\text{z}_1\right)+\arg\left(\text{z}_2\right)$$
