find the principal value of $\log(i^e) ?$

Question

how to find the principal value of $\log(i^e) ?$

My attempts :

$\log \left(e^{{i^e}.\frac{\pi}{2}}\right)$

Now I am not able to proceed further, please help me

Answer 1

Hint:$$\log(i^e)=\log(e^{e\log(i)})=e\log(i)$$ Then use the principal branch of $\log$ to evaluate $\log(i)$. The principal branch is that $$\log(z)=\log|z|+i\arg(z)$$

Answer 2

The principal logarithm ( sometimes capital letters are used to indicate you are working with principal values) of $z$ is $$ \text{Log}\,z=\log|z|+i\, \text{Arg}\,z $$ where $\text{Arg}\,z$ is the principal argument and there will be a convention regarding $\text{Arg}\,z$ e.g. $$ -\pi \lt \text{Arg}\,z \le \pi $$ You can simplify a little to get $$ e\, \text{Log}\, i $$

Then you need to work out $\log|i|$ and $\text{Arg}\, i$.