Let #eta = [3(cos 14^@ + isin 14^@)]^(-4)#

We can see it resembles the trigonometric form of complex numbers :

#z = [r(costheta + isintheta)]^(t)#

We can "mess" around with this identity to furthermore get

#z = r^t[cos(t*theta) + isin(t*theta)]#, derived from de Moivre's theorem.

In our case, #r = 3#, #theta = 14^@# and #t = -4#.

#eta = 1/(3^4)(cos(-56^@) + isin(-56^@))#.

By parity of #sin# and #cos# functions, we have

#eta = 1/(3^4)(cos(56^@) - isin(56^@))#.

Right now, it would've been much nicer if we had #theta = 15^@#, but we'll try to do it anyway.

Both #sin(56^@)# and #cos(56^@)# have "ugly" forms when written in exact value. The best way to do this would be to approximate them with #sin(60^@)# and #cos(60^@)#, respectively.

When doing so, you'll find out that #sin(56^@)# differs from #sin(60^@)# only by about #~ -0.03# and #cos(56^@)# by #~0.05# from #cos(60^@)#.

Plugging these values into our equation, we have :

#eta ~~1/(3^4)(cos(60^@) + 0.05 - i(sin(60^@)-0.03))#

#eta ~~ 1/(3^4)(1/2 + 1/20 - i(sqrt3/2 - 3/100))#

#eta ~~ 11/1620 - i[1/(27sqrt3) - 1/2700]#