How do you evaluate # e^( ( pi)/2 i) - e^( ( 23 pi)/3 i)# using trigonometric functions?

1 Answer
Dwight
Dec 17, 2016

With the help of Euler's formula we can take this right down to
0.5 + #i# (1 + #(sqrt3)/2#)

Explanation:

Start with Euler's formula: #e^(ix)# = #cos x + i Sin x#

Your expression becomes:

cos #pi/2# + #i# sin #pi/2# - cos #(23pi)/3# - #i# sin #(23pi)/3#

To simplify, note cos #pi/2# = 0, sin #pi/2# = 1

and since each complete revolution of a unit circle is 2#pi#, which we will write as #(6pi)/3#, cos #(23pi)/3# = cos #(5pi)/3#
and likewise for sin #(23pi)/3# = sin #(5pi)/3#

So, we now have the original expression simplified down to

#i# - cos #(5pi)/3# - #i# sin #(5pi)/3#

where cos #(5pi)/3# = 0.5 and sin #(5pi)/3# = - #(sqrt3/2)#

So, finally, the result is

0.5 + #i# (1 + #(sqrt3)/2#)

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