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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

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    $\begingroup$ These formulae hold only under suitable constraints/interpretations. The connection is that $\arg = \operatorname{Im}\log$, in the sense that every branch of the argument is the imaginary part of a branch of the logarithm. $\endgroup$ – Daniel Fischer Oct 27 '16 at 9:36
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    $\begingroup$ The general property you're looking for (whereby a map "converts" one operation into another) is called homomorphism: en.wikipedia.org/wiki/Homomorphism $\endgroup$ – Nefertiti Oct 27 '16 at 9:36
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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!

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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 $\text{z}_1\space\wedge\space\text{z}_2\in\mathbb{C}$:

$$\arg\left(\text{z}_1\cdot\text{z}_2\right)=\arg\left(\left|\text{z}_1\right|e^{\arg\left(\text{z}_1\right)i}\cdot\left|\text{z}_2\right|e^{\arg\left(\text{z}_2\right)i}\right)=\arg\left(\left|\text{z}_1\right|\cdot\left|\text{z}_2\right|\cdot e^{\arg\left(\text{z}_1\right)i+\arg\left(\text{z}_2\right)i}\right)=$$ $$\arg\left(\left|\text{z}_1\right|\cdot\left|\text{z}_2\right|\cdot e^{\left(\arg\left(\text{z}_1\right)+\arg\left(\text{z}_2\right)\right)i}\right)=\arg\left(\text{z}_1\right)+\arg\left(\text{z}_2\right)$$

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