How do I use Taylor expansions to show that #e^(itheta) = cos(theta) + isin(theta)#? Using this, what is the value of #e^(ipi) +1#?

1 Answer
Steve M
Nov 23, 2017

# e^(ipi) +1 = 0 #

Explanation:

Firstly as we are seeking Taylor Series pivoted about the origin we are looking at the specific case of MacLaurin Series. Let us start by using the well known Maclaurin Series for the three functions we need:

# \ \ \ \ e^x = 1 +x +(x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + (x^6)/(6!) +... #
# sinx = x - x^3/(3!) + x^5/(5!) - x^7/(7!) + x^9/(9!) -... #
# cosx = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + (x^8)/(8!) - ... #

If we look at the expression #e^(i theta)# we get:

# e^(i theta) = 1 + (i theta) + ((i theta)^2)/(2!) + ((i theta)^3)/(3!) + ((i theta)^4)/(4!) + ((i theta)^5)/(5!) + ((i theta)^6)/(6!) + ... #
# \ \ \ \ = 1 + i theta - (theta^2)/(2!) - (i theta^3)/(3!) + (theta^4)/(4!) + (i theta^5)/(5!) - (theta^6)/(6!) + ... #

# \ \ \ \ = {1 - (theta^2)/(2!) + (theta^4)/(4!) - (theta^6)/(6!) + ... } +#
# \ \ \ \ \ \ \ \ \ \ {i theta- (i theta^3)/(3!)+ (i theta^5)/(5!) + ... } #

# \ \ \ \ = {1 - (theta^2)/(2!) + (theta^4)/(4!) - (theta^6)/(6!) + ... } +#
# \ \ \ \ \ \ \ \ \ \ i{theta- (theta^3)/(3!)+ (theta^5)/(5!) + ... } #

# \ \ \ \ = cos theta + i sin theta \ \ \ # QED

Now if we put #theta =pi# we get:

# e^(ipi) = cos pi + i sin pi #
# \ \ \ \ = -1 + 0 #

Hence,

# e^(ipi) +1 = 0 #

Which is know as Euler's Identity

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