How do you evaluate #e^( ( pi)/12 i) - e^( ( 11 pi)/8 i)# using trigonometric functions?

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Answer 1 · 2016-04-19T06:51:15

#1.139+1.183i#

Euler's formula says that

#e^(ix)=cosx+isinx#.

Using values from the question of #x=pi/12# and #x=(11pi)/8#,

#e^(pi/12i)=cos(pi/12)+isin(pi/12)#
#=cos15+isin15#
#=0.966+0.259i#

#e^((11pi)/8i)=cos((11pi)/8)+isin((11pi)/8)#
#=cos247.5+isin247.5#
#=-0.383-0.924i#

Substituting these values into the question,

#(0.966+0.259i)-(-0.383-0.924i)#
#=0.966+0.383+0.259i+0.924i#
#=1.349+1.183i#