Find the argument of $z = {\left( {2 + i} \right)^{3i}}$ $z = {\left( {2 + i} \right)^{3i}}$
My approach is as follow
$z = {\left( {2 + i} \right)^{3i}} = {\left( {{{\left( {2 + i} \right)}^3}} \right)^i} = {\left( {8 + {i^3} + 12i - 6} \right)^i} = {\left( {2 + 11i} \right)^i}$
$\ln z = i\ln \left( {2 + 11i} \right)$
$2 + 11i = r\cos \theta  + ir\sin \theta  \Rightarrow r = 5\sqrt 5 ;\theta  = {\tan ^{ - 1}}\frac{{11}}{2}$
$ \Rightarrow \ln z = i\ln \left( {2 + 11i} \right) = i\ln \left( {r{e^{i\theta }}} \right) = i\left( {\ln \left( r \right) + \ln \left( {{e^{i\theta }}} \right)} \right) = i\left( {\ln 5\sqrt 5  + i\theta } \right) =  - \theta  + i\left( {\ln 5\sqrt 5 } \right)$
$  \Rightarrow z = {e^{\left( { - \theta  + i\left( {\ln 5\sqrt 5 } \right)} \right)}} = {e^{ - \theta }}{e^{i\left( {\ln 5\sqrt 5 } \right)}} \Rightarrow r{e^{i\phi }}$
The modulus is $r = {e^{ - {{\tan }^{ - 1}}\frac{{11}}{2}}}$, Argument $= \phi  = \ln 5\sqrt 5 $
How do we convert this argument into angle

$$Z=(2+i)^{3i} \implies \ln Z=3i \ln (2+i)=3 i[\ln \sqrt{5}+i \tan^{-1}\frac{1}{2}]$$
$$\implies \ln Z= -3\tan^{-1}(1/2)+3i\ln \sqrt{5} \implies Z=e^{-3\cot^{-1}2}~~ e^{3i \ln \sqrt{5}}$$
$$\implies \arg(Z)=3\ln\sqrt{5}=2.414<\pi$$
AS this lies in $(-\pi, \pi]$, it is also the principal value of the argument.
Here we have used $\ln z=\ln\sqrt{x^2+y^2}+i\tan^{-1}(y/x), \arg (r e^{i\theta})=\theta$
