How can you use trigonometric functions to simplify # 23 e^( ( pi)/8 i ) # into a non-exponential complex number?

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Answer 1 · 2016-04-08T15:11:50

#23e^(pi/8i) = - 20.08 - 11.2i#

According to Euler's formula,

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

When #pi/8# is substituted as #x#, then

#e^(pi/8i) = cos(pi/8) + isin(pi/8)#
# = cos(22.5^o) + isin(22.5^o)#
# = -0.873 -0.487i#,

which is the value of #e^(pi/8i)#.

Multiply the whole thing by #23# to find the entire thing:

#23(- 0.873 - 0.487i) = - 20.08 - 11.2i#