Evaluate $(1+i)^{(1-2i)}$ Find all the values of $(1+i)^{(1-2i)}$ and show that there are small values as we wish (else from $0$) and big values as we wish
\begin{align} 
(1+i)^{(1-2i)}&=e^{\ln(1+i)^{(1-2i)}}
=e^{(1-2i)\ln(1+i)}
\\&=e^{(\ln\sqrt{2}+i(\frac{\pi}{2}+2\pi k))(1-2i)}
\\&=e^{\ln\sqrt{2}-2\ln\sqrt{2}*i+i(\frac{\pi}{2}+2\pi k)+2(\frac{\pi}{2}+2\pi k))}
\\&=\sqrt{2}*e^{-2\ln\sqrt{2}*i}*i*e^{2(\frac{\pi}{2}+2\pi k)}
\\&=e^{-2\ln\sqrt{2}*i+\pi+4\pi k}+i\sqrt{2}
\\&=e^{-i(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=e^{-i}*e^{(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=\frac{3}{4}*e^{(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=\frac{3}{4}*2e^{i(\pi+4\pi k)}+i\sqrt{2}
\\&=\frac{-3}{2}+i\sqrt{2}
\end{align}
What can I conclude from it on the values?
Or should I stop here $e^{-2\ln\sqrt{2}*i+\pi+4\pi k}+i\sqrt{2}$ and $e^{\pi+4 \pi k}cis(-2\ln\sqrt{2}+\frac{\pi}{2})$

Hint. Note that $|e^z|=|e^{\text{Re}(z)+i\text{Im}(z)}|=e^{\text{Re}(z)}$, hence here we are interested in (see your third step where the correct argument of $(1+i)$ is $\pi/4+2\pi k$)
$$\text{Re}(\ln\sqrt{2}+i(\frac{\pi}{4}+2\pi k))(1-2i))=\ln\sqrt{2}+\pi\left(4k+\frac{1}{2}\right)$$
where $k\in\mathbb{Z}$. Now consider the limits as $k\to \pm \infty$.
