How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( pi)/6 i)# using trigonometric functions?

1 Answer
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Jan 31, 2016

#e^{i (7pi)/4} - e^{i pi/6} = (sqrt2-sqrt3)/2 - i ((1+sqrt2)/2)#

Explanation:

Use the Euler's Formula, which states that #e^{i theta} -= cos(theta) + i sin(theta)#. (Proof omitted)

Therefore,

#e^{i (7pi)/4} - e^{i pi/6} = (cos((7pi)/4) + i sin((7pi)/4)) - (cos(pi/6) + i sin(pi/6))#

#= (sqrt2/2 + i (-sqrt2/2)) - (sqrt3/2 + i (1/2))#

#= (sqrt2-sqrt3)/2 - i ((1+sqrt2)/2)#

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