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

Final Answer

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