How to evaluate $$\log(i)?$$Also, how to evaluate $$\log(3+4i)?$$I am reading complex analysis and I know that logarithm is a multibranched function and is periodic. What I have for the definition is $$\log(z) = \log(\rho)+i(\theta +2k\pi).$$That is for when $z$ is complex. I also have, $\log(z)=\log|z|+i(\arg(z)+2k\pi)$I don't know where to start, so please write a detailed and step-by-step solution if you can, thank you very much!
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$\begingroup$The idea is that:
$$\log(z) = \log(\rho)+i(\theta +2k\pi),$$
where $z = \rho e^{i\theta}.$
When $z = i$, then $z = e^{i\frac{pi}{2}}$. Hence, $\rho = 1$ and $\theta = \frac{\pi}{2}$. Finally:
$$\log(i) = \log(1) + i \left(\frac{\pi}{2} + 2k \pi\right) = \frac{i\pi}{2}(1 + 4k). $$
When $z = 3+4i$, we have that $\rho = \sqrt{3^2 + 4^2} = 5$ and $\theta = \arctan \left(\frac{4}{3}\right).$
Hence:
$$\log(3+4i) = \log(5) + i \left(\arctan \left(\frac{4}{3}\right) + 2k \pi\right). $$
$\endgroup$ $\begingroup$The polar form $a+ib$ ($a$ and $b$ real) is $re^{i\theta}$ where $r= \sqrt {a^{2}+b^{2}}$ and $\theta =\arctan \frac b a$ (which is taken as $\pi /2$ if $a=0$ and $b >0$, as $-\pi /2$ if $a=0$ and $b <0$).
This gives $i=e^{i\pi /2}$ and $3+4i=5e^{i\arctan (4/3)}$. Now from the definition of logarithm we get $\log i=i\pi /2+2k\pi i$ and $\log (3+4i)= \ln 5+i\arctan (\frac 4 3) +2k \pi i$.
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