e^((23pi)i/12) = cos(23pi/12) + isin(23pi/12)e(23π)i12=cos(23π12)+isin(23π12)
Since 23pi/12 = 2pi - pi/1223π12=2π−π12 so we have
e^((23pi)i/12) = cos(2pi - pi/12) + isin(2pi - pi/12)e(23π)i12=cos(2π−π12)+isin(2π−π12)
But since 2pi2π is the period, we can safely ignore it
e^((23pi)i/12) = cos(-pi/12) + isin(-pi/12)e(23π)i12=cos(−π12)+isin(−π12)
Remember that cos(-theta) = cos(theta)cos(−θ)=cos(θ) and sin(-theta) = -sin(theta)sin(−θ)=−sin(θ)
e^((23pi)i/12) = cos(pi/12) - isin(pi/12)e(23π)i12=cos(π12)−isin(π12)
From there, it's using half angle formulas to figure out these values
sin(u/2) = sqrt((1-cos(u))/2)sin(u2)=√1−cos(u)2 and cos(u) = sqrt((1+cos(u))/2)cos(u)=√1+cos(u)2
So
sin(pi/12) = sqrt((1 - sqrt(3)/2)/2) = sqrt((2-sqrt(3))/4) = sqrt(2-sqrt(3))/2sin(π12)=√1−√322=√2−√34=√2−√32
cos(pi/12) = sqrt((1 + sqrt(3)/2)/2) = sqrt((2+sqrt(3))/4) = sqrt(2+sqrt(3))/2cos(π12)=√1+√322=√2+√34=√2+√32
e^((23pi)i/12) = sqrt(2+sqrt(3))/2 -isqrt(2-sqrt(3))/2e(23π)i12=√2+√32−i√2−√32
And lastly, multiply by 1616 so we have
z = 8sqrt(2+sqrt(3)) -8sqrt(2-sqrt(3))iz=8√2+√3−8√2−√3i
