Stackexchange community,
i have a question concerning the rules for exponentials and logarithms with complex arguments.
For real arguments we have:
$$e^{a+b}=e^a\cdot e^b$$ $$e^{a-b}=e^{a}/e^{b}$$ $$e^{\ln a}=a$$ $$\ln(a\cdot b)=\ln a + \ln b$$ $$\ln(\frac{a}{b})=\ln a - \ln b$$ $$\ln(e^{a})=a$$ $$\log_a(b)=\frac{\log_c(b)}{\log_c(a)}$$
Can i use these laws also for complex numbers? I know I have to take care of periodicity and singularities.
For exponential, it works the same. For the logarithme it is a little bit different. $$\ln(ab)=\ln(|a||b|)+i\arg(ab)=\ln(|a|)+\ln(|b|)+i\arg(ab)$$
but $$\arg(ab)\equiv \arg(a)+\arg(b)\pmod {2\pi},$$ therefore $$\ln(ab)=\ln(a)+\ln(b)\iff \arg(a)+\arg(b)\in]-\pi,\pi[,$$ otherwise $$\ln(ab)=\ln(a)+\ln(b)+2ik\pi$$ if $\arg(a)+\arg(b)\in]-\pi+2k\pi,\pi+2k\pi[$.
Finally, if you use the property I mentionned before, you will get $$\ln(e^z)=z\iff \Im(z)\in ]-\pi,\pi[.$$
Consider for instance the first identity, $e^{a+b} = e^ae^b$. To show that this holds for all complex numbers $a$ and $b$ using that you know it holds for real numbers $a$ and $b$, you could argue as follows. Suppose first that $b$ is real, and let $f(z) = e^{z+b}$ and $g(z) = e^ze^b$. These are two holomorphic functions which agree whenever $z$ is real, and therefore by the principle of analytic continuation $f = g$ everywhere. Thus, assuming $b$ is real, $e^{z+b} = e^ze^b$ for all complex numbers $z$. We can now argue like this one more time: now let $a$ be any complex number, and let $f(w) = e^{a+w}$ and $g(w) = e^ae^w$. By the argument just made, we know that $f(w) = g(w)$ whenever $w$ is real, and therefore by analytic continuation $f = g$ everywhere. This proves $e^{a+b} = e^ae^b$ for all complex numbers $a$ and $b$.
Using arguments like these, you can transfer identities you know from the real case to the complex case. As you say, though, when dealing with the logarithms, you really must be very careful to take into account the fact that logarithms are multi-valued functions.
