How do you evaluate # e^( ( 7 pi)/6 i) - e^( ( 14 pi)/3 i)# using trigonometric functions?

1 Answer
Mar 30, 2016

#-(sqrt3 -1)/2 - i (sqrt3 +1)/2#

Explanation:

#e^((7pi i)/6)= cos ((7pi)/6) + i sin ((7pi)/6)#

= #-cos pi/6 - i sin pi/6# #[(7pi)/6 = pi +pi/6#, In the third quadrant both sin and cos would b negative]

Like wise #e^((14pi i)/3)= cos ((14pi)/3) + i sin ((14pi)/3)#

=# cos ((2pi)/3) + i sin ((2pi)/3)# # [(14pi)/3= 4pi +(2pi)/3= (2pi)/3]#

=# - cos (pi/3) +i sin (pi/3) #

Now combining both expressions, it would be #-sqrt3 /2 -i/2 +1/2 -i sqrt3 /2#