Yes. Write a complex number $a+bi$ in polar form $ce^{it}$. Then $$a^2+b^2=(a+bi)(a-bi)=(ce^{it})(ce^{-it})=c^2.$$ This is the first theorem I prove in my complex analysis class (after defining complex multiplication via the polar form and checking that it agrees with the more traditional definition via $i^2=-1$ and distributivity).

Yes. Write a complex number $a+bi$ in polar form $ce^{it}$. Then $$a^2+b^2=(a+bi)(a-bi)=(ce^{it})(ce^{-it})=c^2.$$ This is the first theorem I prove in my complex analysis class (after defining complex multiplication via the polar form and checking that it agrees with the more traditional definition via $i^2=-1$ and distributivity).

Added. See the comments for more details what this proof is really based on, and why it is not circular.