Editing Euler's formula (section)

==Proofs==
Various proofs of the formula are possible.

===Using differentiation===
This proof shows that the quotient of the trigonometric and exponential expressions is the constant function one, so they must be equal (the exponential function is never zero,<ref name=Apostol>{{cite book |last=Apostol |first=Tom |title=Mathematical Analysis |page=20 |publisher=Pearson |date=1974 |isbn=978-0201002881}} Theorem 1.42</ref> so this is permitted).<ref>user02138 (https://math.stackexchange.com/users/2720/user02138), How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?, URL (version: 2018-06-25): https://math.stackexchange.com/q/8612</ref>

Consider the function {{math|''f''(''θ'')}} 
<math display="block"> f(\theta) = \frac{\cos\theta + i\sin\theta}{e^{i\theta}} = e^{-i\theta} \left(\cos\theta + i \sin\theta\right) </math>
for real {{mvar|θ}}. Differentiating gives by the [[product rule]]
<math display="block"> f'(\theta) = e^{-i\theta} \left(i\cos\theta - \sin\theta\right) - ie^{-i\theta} \left(\cos\theta + i\sin\theta\right) = 0</math>
Thus, {{math|''f''(''θ'')}} is a constant. Since {{math|1=''f''(0) = 1}}, then {{math|1=''f''(''θ'') = 1}} for all real {{mvar|θ}}, and thus 
<math display="block">e^{i\theta} = \cos\theta + i\sin\theta.</math>

===Using power series===
[[File:Eulers-forrmula-standalone.svg|thumb|alt=Each successive term in the series rotates 90 degrees counter clockwise. The even-power terms are real, hence parallel to the real line, and the odd-power terms are imaginary, hence parallel to the imaginary axis. Plotting each term as a vectors in the complex plane lying end-to-end (vector addition) results in a piecewise-linear spiral starting from the origin and converging to the point (cos 2, sin 2) on the unit circle. |A plot of the first few terms of the Taylor series of {{math|''e''<sup>''it''</sup>}} for {{math|''t'' {{=}} 2}}. ]]

Here is a proof of Euler's formula using [[Taylor series|power-series expansions]], as well as basic facts about the powers of {{mvar|i}}:<ref>{{cite book|url=https://books.google.com/books?id=PjK0F0T3NBoC&pg=PA428 |title=A Modern Introduction to Differential Equations |first=Henry J. |last=Ricardo |date=23 March 2016 |page=428|publisher=Elsevier Science |isbn=9780123859136 }}</ref>
<math display="block">\begin{align}
i^0 &= 1, & i^1 &= i, & i^2 &= -1, & i^3 &= -i, \\
i^4 &= 1, & i^5 &= i, & i^6 &= -1, & i^7 &= -i \\
&\vdots   & &\vdots   & &\vdots    & &\vdots
\end{align}</math>

Using now the power-series definition from above, we see that for real values of {{mvar|x}}
<math display="block">\begin{align}
 e^{ix} &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \frac{(ix)^8}{8!} + \cdots \\[8pt]
 &= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}{8!} + \cdots \\[8pt]
 &= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) \\[8pt]
 &= \cos x + i\sin x ,
\end{align}</math>
where in the last step we recognize the two terms are the [[Taylor series#Trigonometric functions|Maclaurin series]] for {{math|cos ''x''}} and {{math|sin ''x''}}. [[Riemann series theorem|The rearrangement of terms is justified]] because each series is [[absolute convergence|absolutely convergent]].

===Using polar coordinates===
Another proof<ref name=Strang>{{cite book |url=http://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/textbook/ |title=Calculus |first=Gilbert |last=Strang |page=389 |publisher=Wellesley-Cambridge |year=1991 |isbn=0-9614088-2-0}} Second proof on page.</ref> is based on the fact that all complex numbers can be expressed in [[polar coordinates]]. Therefore, [[for some]] {{mvar|r}} and {{mvar|θ}} depending on {{mvar|x}},
<math display="block">e^{i x} = r \left(\cos \theta + i \sin \theta\right).</math>
No assumptions are being made about {{mvar|r}} and {{mvar|θ}}; they will be determined in the course of the proof. From any of the definitions of the exponential function it can be shown that the derivative of {{math|''e''<sup>''ix''</sup>}} is {{math|''ie''<sup>''ix''</sup>}}. Therefore, differentiating both sides gives
<math display="block">i e ^{ix} = \left(\cos \theta  + i \sin \theta\right) \frac{dr}{dx} + r \left(-\sin \theta + i \cos \theta\right) \frac{d\theta}{dx}.</math>
Substituting {{math|''r''(cos ''θ'' + ''i'' sin ''θ'')}} for {{math|''e<sup>ix</sup>''}} and equating real and imaginary parts in this formula gives {{math|1=''{{sfrac|dr|dx}}'' = 0}} and {{math|1=''{{sfrac|dθ|dx}}'' = 1}}. Thus, {{mvar|r}} is a constant, and {{mvar|θ}} is {{math|''x'' + ''C''}} for some constant {{mvar|C}}. The initial values {{math|1=''r''(0) = 1}} and {{math|1=''θ''(0) = 0}} come from {{math|1=''e''<sup>0''i''</sup> = 1}}, giving {{math|1=''r'' = 1}} and {{math|1=''θ'' = ''x''}}. This proves the formula
<math display="block">e^{i \theta} = 1(\cos \theta +i  \sin \theta) = \cos \theta + i  \sin \theta.</math>