Vtyjkyuk

Find the value of

$$left(dfracsqrt3+i2right)^69.DeclareMathOperatorciscis$$

I tried to solve this complex expression by converting it into polar form.
I expressed it in polar form $rcis(t)$ from rectangular form $x+iy$ where $cis(t) = cos(t) + isin(t)$.
But I am unable to solve further due to the exponent of 69!

$$left(dfracsqrt3+i2right)^69=left(dfracsqrt32+dfrac12iright)^69=(cosdfracpi6+isindfracpi6)^69=cosdfrac69pi6+isindfrac69pi6=-i$$
by De Moivre's formula.

Converting to polar coordinates is probably the best way to go.

$$r = sqrtleft(fracsqrt32right)^2+left(frac12right)^2=1$$

$$theta = arctan(1/sqrt3)=fracpi6$$

Then $$(re^theta i )^69=e^frac69pi6i=e^10pi i+frac3pi2i=e^frac3pi2i=-i$$

$$fracsqrt3+i2=e^ifracpi6$$

$$big(fracsqrt3+i2big)^69=e^ifracpi6cdot69=e^i(11pi+fracpi2)=e^i(12pi-fracpi2)=e^-ifracpi2=-i$$

I used the fact that $e^i(2npi)=1$ where $nin Z$

Bruteforcing:
$$beginalignleft(dfracsqrt3+i2right)^69&=left[left(dfracsqrt3+i2right)^2right]^34cdot fracsqrt3+i2=\
&=left[left(frac1+sqrt3i2right)^2right]^17cdot fracsqrt3+i2=\
&=left[left(frac-1+sqrt3i2right)^2right]^8cdot frac-1+sqrt3i2cdot fracsqrt3+i2=\
&=left[left(frac-1-sqrt3i2right)^2right]^4cdot frac-sqrt3+i2=\
&=left[left(frac-1+sqrt3i2right)^2right]^2cdot frac-sqrt3+i2=\
&=left[frac-1-sqrt3i2right]^2cdot frac-sqrt3+i2=\
&=frac-1+sqrt3i2cdot frac-sqrt3+i2=\
&=-i.endalign$$
Alternatively:
$$E=left(fracsqrt3+i2right)^3cdot 23=left(frac3sqrt3+9i-3sqrt3-i8right)^23=i^23=(i^2)^11cdot i=-i.$$