Just had a quick question in the solution.
The complex number somehow goes from a power of 6 to simple solution at the end. I was wondering is there a quick way I am missing out on? Or will one convert it to Polar and then apply DeMoivre's Theorem.
Thanks for your time
One efficient approach is to convert $1+i2$ to polar coordiates $\sqrt{5}e^{i\arctan(2)}$. Then, $$(1+i2)^6=125e^{i6\arctan(2)}=117+i44$$
Alternatively and less efficiently, we could use the binomial theorem and write
$$\begin{align} (1+i2)^6&=\sum_{k=0}^6\binom{6}{k}(i2)^k\\\\ &=\binom{6}{0}+\binom{6}{1}(i2)+\binom{6}{2}(i2)^2+\binom{6}{3}(i2)^3+\binom{6}{4}(i2)^4+\binom{6}{5}(i2)^5+\binom{6}{6}(i2)^6\\\\ &=(1-60+240-64)+i(12-160+192)\\\\ &=117+i44 \end{align}$$