How do you evaluate # e^( ( pi)/4 i) - e^( ( 2 pi)/3 i)# using trigonometric functions?

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Answer 1 · 2016-07-02T20:24:56

Recall Euler's formula:

#e^(itheta) = costheta + isintheta#

Using this:

#e^((ipi)/4) - e^((2ipi)/3) = cos(pi/4) + isin(pi/4) - cos((2pi)/3) - isin((2pi)/3)#

Evaluating these trig functions we get:

#(sqrt(2))/2 + i(sqrt(2))/2 + 1/2 -i(sqrt(3))/2#

And grouping terms together gives us

# (sqrt(2)+1)/2 + i((sqrt(2) - sqrt(3))/2)#