How do you evaluate # e^( ( pi)/4 i) - e^( ( 11 pi)/6 i)# using trigonometric functions?

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Answer 1 · 2017-05-26T02:16:45

#e^(pi/4i)-e^((11pi)/6i)=(sqrt2-sqrt3)/2 + (sqrt2+1)/2i#

There is a special formula for complex numbers that we will use to solve this problem:

#e^(i theta) = costheta +isintheta#

Therefore, we can rewrite this problem as:

#e^(pi/4i)-e^((11pi)/6i)#

#=cos(pi/4)+isin(pi/4)-cos((11pi)/6)-isin((11pi)/6)#

Now all we have to do is simplify.

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

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

So #e^(pi/4i)-e^((11pi)/6i)=(sqrt2-sqrt3)/2 + (sqrt2+1)/2i#

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