How do you evaluate # e^( ( 19 pi)/12 i) - e^( ( pi)/4 i)# using trigonometric functions?

1 Answer
Shwetank Mauria
May 19, 2016

#e^((19pi)/12)-e^(pi/4)=-0.4483-i1.673#

Explanation:

#e^((19pi)/12i)=cos((19pi)/12)+isin((19pi)/12)# and #e^(pi/4i)=cos(pi/4)+isin(pi/4)#

Hence, #e^((19pi)/12)-e^(pi/4)=cos((19pi)/12)+isin((19pi)/12)-cos(pi/4)-isin(pi/4)#

= #cos((-5pi)/12)-cos(pi/4)+i(sin((-5pi)/12)-sin(pi/4))#

= #cos((5pi)/12)-1/sqrt2+i(-sin((5pi)/12)-1/sqrt2)#

= #0.2588-0.7071+i(-0.9659-0.7071)#

= #-0.4483-i1.673#

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