How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( 2 pi)/3 i)# using trigonometric functions?

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Answer 1 · 2017-08-20T03:36:34

Use Euler's identity...

...which I will use, but not prove, here:

#e^(ix) = cosx + isinx#

so, for the first term, we have #x = (7pi)/4#
so the first term can be re-written:

#cos((7pi)/4) + isin((7pi)/4)#

and the second can be rewritten:

#cos((2pi)/3) + isin((2pi)/3)#

So, plugging it all back in:

#(cos((7pi)/4) + isin((7pi)/4)) - (cos((2pi)/3) + isin((2pi)/3))#

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