From part (c) we have the polar form z = − 1 / 2 + i 3 / 2 = e 2 3 π i {\displaystyle z=-1/2+i{\sqrt {3}}/2=e^{{\frac {2}{3}}\pi i}} , so
z 23 = ( e 2 3 π i ) 23 = e 2 3 ⋅ 23 π i = e 46 3 π i = e ( 14 π i + 4 3 π i ) = e 14 π i e 4 3 π i . {\displaystyle z^{23}=(e^{{\frac {2}{3}}\pi i})^{23}=e^{{\frac {2}{3}}\cdot 23\pi i}=e^{{\frac {46}{3}}\pi i}=e^{(14\pi i+{\frac {4}{3}}\pi i)}=e^{14\pi i}e^{{\frac {4}{3}}\pi i}.}
Notice that e 14 π i = e 2 π i ⋅ 7 = ( e 2 π i ) 7 = 1 7 = 1 , {\displaystyle e^{14\pi i}=e^{2\pi i\cdot 7}=(e^{2\pi i})^{7}=1^{7}=1,}
since e 2 π i = cos ( 2 π ) + i sin ( 2 π ) = 1 + 0 i = 1. {\displaystyle e^{2\pi i}=\cos(2\pi )+i\sin(2\pi )=1+0i=1.}
So z 23 = e 4 3 π i = cos ( 4 3 π ) + i sin ( 4 3 π ) = − 1 2 − i 3 2 . {\displaystyle z^{23}=e^{{\frac {4}{3}}\pi i}=\cos \left({\frac {4}{3}}\pi \right)+i\sin \left({\frac {4}{3}}\pi \right)=\color {blue}-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}.}
, so
