How do you evaluate # e^( ( 5 pi)/4 i) - e^( ( 11 pi)/4 i)# using trigonometric functions?

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Answer 1 · 2018-08-01T15:24:37

#= sqrt2 i#

Use #e^(i pi ) = - 1 and e^(i 2 pi) = 1#

#e^(i5/4pi) - e^(i11/4pi)#

#= e^(ipi/4)e^(i pi) - e^(- i pi/4)e^(i 2 pi )#

#= e^(ipi/4)(-1) - e^(- i pi/4) ( 1 )#

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

#= 2 sin ( pi/4 ) i#

# sqrt2 i#