I do believe this is a simple question, but I am still unsure if my solution correct.
We have that: $$z^\lambda=e^{\ln(z)\cdot \lambda}$$ And also that: $$\ln(z)=\ln\left(\sqrt{a^2+b^2}\right) + i\cdot \arctan \left(\frac{b}{a}\right),$$ where $a$ and $b$ are the real and imaginary parts of $z$. Thus, with that in mind: $1^i=e^{\ln(1)\cdot i}$.
However, $$\ln(1)=\ln(1)+i \cdot \arctan(0)\Rightarrow \ln(1)=i\cdot k\pi; \ k=-1,0,1.$$ Finally, we can say that: $$1^i=e^{ik\pi\cdot i}\Rightarrow 1^i=e^{-k\pi}; \ k=-1,0,1$$ Using the main branch, that is the solution.
Now, is my solution right or wrong? As a beginner in such topic, I don’t know for certain. If wrong, what exactly did I do wrong? Thank you for your attention!
The main branch refers to the choice of $k$.
Basically the problem boils down to the fact that arguments repeat every $2\pi$ hence $$1=e^{0i}=e^{2\pi i}=e^{4\pi i}=e^{2k\pi i }\quad k\in\mathbb{Z}.$$
Being in the main branch means that the arguments should stay in the range $(-\pi,\pi)$ hence in this case $k=0$, i.e., $1=e^{0i}$.
From this you get $$1^i=(e^{0i})^i=e^{0i^2}=e^{-0}=1.$$
If instead you were to choose another branch such as $(\pi,3\pi)$ then $k=1$ and hence $1=e^{2\pi i}$ so that $$1^i=(e^{2\pi i})^i=e^{2\pi i^2}=e^{-2i}.$$