<s>[INST] 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
 [/INST] Another possible approach:
Taking principal logarithm both side of $z=(2+i)^{3i}$,
$Log z=3i[\ln \sqrt{2^2+1^2}+i ~tan^{-1}(1/2)]=3i[\ln \sqrt 5+i ~tan^{-1}(1/2)]$.
Thus $Log z=-3 tan^{-}(1/2)+i~3 \ln \sqrt 5$ and so $z=e^{-3tan^{-1}(1/2)} \cdot e^{i3 \ln \sqrt 5}$.
This gives us $z=e^{-3tan^{-1}(1/2)} (cos(3 \ln \sqrt 5)+i ~sin (3 \ln \sqrt 5))$.
Thus, $\text{Arg}(z)=tan^{-1}(tan(3 \ln \sqrt{5}))=3 \ln \sqrt 5$, because $\forall x \in (-\pi/2,~\pi/2), \ tan^{-1}(tan ~x)=x.$
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