How can you use trigonometric functions to simplify # 16 e^( ( pi)/6 i ) # into a non-exponential complex number?

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Answer 1 · 2015-12-31T10:23:42

Euler formula #e^(i theta) = cos(theta)+isin(theta)# using this we can represent the given complex number as .

#16e^(pi/6i) = 8sqrt(3)+8i#

#16e^(pi/6i)#
Comparing to the Euler's formula #e^(i theta)#
We can see #theta=pi/6#

Therefore,

#16e^(pi/6 i) = 16(cos(pi/6) + isin(pi/6))#

#16e^(pi/6i) = 16(sqrt(3/2) + i (1/2))#

#16e^(pi/6i) = 16/2(sqrt(3) + i)#

#16e^(pi/6i) = 8(sqrt(3)+i)#

#16e^(pi/6i) = 8sqrt(3)+8i# Answer