How can I calculate $\log(e^{e^i})$?

Question

I'm studying complex analysis and I'm wondering how to calculate the following multivalued function (using the expression $\log(z)=\ln|z| + i\operatorname{Arg(z)}$):

$$\Large \log(e^{e^i})$$

Thank you!

Answer 1

The most general logarithm would be $e^i+2n\pi i=\cos 1+(\sin 1+2n\pi)i$ with $n\in\Bbb Z$, as @AsemAbdelraouf noted. The principal natural logarithm is the case $n=0$ (as @LeonhardEuler discussed), with phase $\in(-\pi,\,\pi]$.

Answer 2

The exponential function $e^w$, with $w=a+ib$, is defined as $e^{w}=e^{a}\cos b+ie^{a}\sin b$. So, for $w=i$, $a=b=1$, we have the particular value of the Euler's formula $e^{i}=\cos \left( 1 \right) +i\sin \left( 1 \right)$.

Now let $z=e^{e^{i}}$. Since $e^{w\pm i2k\pi}=e^{w}e^{\pm i2k\pi}=e^{w}\times 1=e^{w}$, we then have

\begin{equation*} z=e^{e^{i}}=e^{\cos \left( 1\pm 2k\pi \right) }e^{i\sin \left( 1\pm 2k\pi \right) } \end{equation*}

and

$$ \left\vert z\right\vert =\left\vert e^{\cos \left( 1\pm 2k\pi \right) }\right\vert \left\vert e^{i\sin \left( 1\pm 2k\pi \right) }\right\vert =\left\vert e^{\cos \left( 1\pm 2k\pi \right) }\right\vert\times 1 =e^{\cos \left( 1\right) }. $$

The principal argument of $z$, $\operatorname{Arg}(z)$, is such that $-\pi <\operatorname{Arg}(z)\leq \pi $. So $\operatorname{Arg}(z)=\sin \left( 1\right) $ occurs for $k=0$, and, by definition, the principal value of $\log (z)$, $\operatorname{Log}(z)$, is equal to

\begin{equation*} \operatorname{Log}(z)=\log(\left\vert z\right\vert) +i\operatorname{Arg}(z)=e^{\cos \left( 1\right) }+i\sin \left( 1\right) , \end{equation*}

where $\log(|z|)$ denotes the natural logarithm of $|z|$.