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===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>
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