How do you evaluate # e^( ( 3 pi)/8 i) - e^( ( 19 pi)/6 i)# using trigonometric functions?

1 Answer
reudhreghs
Apr 9, 2016

#1.249 + 1.424i#

Explanation:

According to Euler's formula,

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

Using values for #x# from the question,

#x = (3pi)/8#
#e^((3pi)/8i) = cos((3pi)/8) + isin((3pi)/8)#
# = cos67.5 + isin67.5#
# = 0.383 + 0.924i#

#x = (19pi)/6#
#e^((19pi)/6i) = cos((19pi)/6) + isin((19pi)/6)#
# = cos570 + isin570#
# = -0.866 - 0.500i#

Putting these two values together,

#0.383 + 0.924i + 0.866 + 0.500i = 1.249 + 1.424i#

You can check the trigonometric values, there's a margin for error with rounding (or if I've accidentally done radians rather than degrees or vice versa on a calculator) but the equation is correct. Simply use Euler's formula, enter your values for #x# and solve with a calculator.

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