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The question is how can I solve $$(1+2i)^i$$ Thanks for hints.

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    Please add a bit more context. Such as your own effort. You may wish to consider some similar things, such as $$i^{i}=e^{-\frac{\pi}{2}}$$ – 2017-02-03
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    How did you try to solve it? – 2017-02-03
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    Put $1 + 2i$ into polar coordinates. $1 + 2i = r(\cos \theta + i \sin \theta) = e^{r + \theta i}$ then $(1+2i)^i = e^{-\theta + ri} = \frac 1{e^{\theta}}(\cos r + i \sin r)$. – 2017-02-03
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    I don't think "solve" is the right word here?... maybe "expand" or "simplify" or ... – 2017-02-03

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$(1+2i) = \sqrt 5 (\cos\theta + i \sin \theta)$ where $\theta = \arctan 2$

now for the tricky part!

$\sqrt 5 (\cos\theta + i \sin \theta) = e^{\sqrt 5 + i\theta}$

This is Euler's identity.

$(\sqrt 5 (\cos\theta + i \sin \theta))^i = e^{(\sqrt 5 + i\theta)^i} =e^{(-\theta + i \sqrt 5)} = e^{-\arctan 2}(\cos \sqrt 5 + i \sin \sqrt 5) $