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$z = {\left( {2 + i} \right)^{3i}}$

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

halrankard
  • 3,454
Samar Imam Zaidi
  • 8,800

2 Answers2

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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 $\forall x \in (-\pi/2,~\pi/2), \ tan^{-1}(tan ~x)=x.$

MAS
  • 10,638
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$$Z=(2+i)^{3i} \implies \ln Z=3i \ln (2+i)=3 i[\ln \sqrt{5}+i \tan^{-1}\frac{1}{2}]$$ $$\implies \ln Z= -3\tan^{-1}(1/2)+3i\ln \sqrt{5} \implies Z=e^{-3\cot^{-1}2}~~ e^{3i \ln \sqrt{5}}$$ $$\implies \arg(Z)=3\ln\sqrt{5}=2.414<\pi$$ AS this lies in $(-\pi, \pi]$, it is also the principal value of the argument. Here we have used $\ln z=\ln\sqrt{x^2+y^2}+i\tan^{-1}(y/x), \arg (r e^{i\theta})=\theta$

Z Ahmed
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