# Euler's Identity

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I couldn't help but wonder if anyone knows how raising e to the imaginary pi power could result in a negative number.  In this case, -1.

Different values of pi can result in different numbers, like 2pi would give a positive one.  Then it seems that raising e by a multiple of imaginary pi powers can result in a positive or negative answer.  For example, you wouldn't ever be able to raise 1 to an imaginary pi power to ever get a negative answer.  You would always just get a positive one.  How is e so much more special than 1 to where it's imaginary pi powers can be negative.  Then you can raise 2 to an imaginary power of pi and get a negative complex number.

Is there some rule of raising things to imaginary powers that I do not know about that allows the answer to become negative?

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3 hours ago, Conjurer said:

Is there some rule of raising things to imaginary powers that I do not know about that allows the answer to become negative?

I'm really bad at math, but the trigonometric rotation?

Here is what I found. To check.

Strictly positive reals have an argument that is multiple of 2π, strictly negative reals have an odd multiple of π as argument. Pure non-zero imaginaries have an argument congruent to π/2 or –π/2 modulo 2π, depending on the sign of their imaginary part. The particular values, which are the intersections of the trigonometric circle with the real and imaginary axes, are:

$$\operatorname {e} ^{\mathrm {i} 0}=\operatorname {e} ^{0}=1}\\ \operatorname {e} ^{\mathrm {i} \pi }=\cos \pi +\mathrm {i} \sin \pi =-1}\\ \operatorname {e} ^{\mathrm {i} \pi /2}=\cos(\pi /2)+\mathrm {i} \sin(\pi /2)=\mathrm {i} }\\ \operatorname {e} ^{\mathrm {i} (-\pi /2)}=\cos(-\pi /2)+\mathrm {i} \sin(-\pi /2)=-\mathrm {i} }\\ \operatorname {e} ^{\mathrm {-} i(\pi /2)}=\cos(\pi /2)-\mathrm {i} \sin(\pi /2)=-\mathrm {i} }\$$

Edited by Kartazion
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Nice video explaining the whole thing:

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4 hours ago, Conjurer said:

Different values of pi can result in different numbers, like 2pi would give a positive one.  Then it seems that raising e by a multiple of imaginary pi powers can result in a positive or negative answer.

It's not magic, just simple algebra

${e^{2i\pi }} = {\left( {{e^{i\pi }}} \right)^2} = \left( {{e^{i\pi }}} \right)\left( {{e^{i\pi }}} \right) = \left( { - 1} \right)\left( { - 1} \right) = 1$

Does this help ?

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9 hours ago, Kartazion said:

Here is what I found. To check.

ei0=e0=1eiπ=cosπ+isinπ=1eiπ/2=cos(π/2)+isin(π/2)=iei(π/2)=cos(π/2)+isin(π/2)=iei(π/2)=cos(π/2)isin(π/2)=i

I think I found the answer on an old university web page.

It appears that Euler's Equation actually defines raising something to a complex power.  Then to raise a^(b + ic) = a^b(cos(c ln(a)) + i sin(c ln(a))).

It turns out that the natural log of 1 is zero, so then the imaginary sin portion reduces to zero for all powers of 1.  Then the cos of zero reduces to 1.  Then raising 1 to any power just always gets multiplied by 1 when it is raised to a complex power.  Then 1 to any power of b is still just one.  I guess it checks out that one to any power is still just one, even if that power is imaginary.

Surprisingly, I have not been able to find any holes in the Euler formula {e^(ix) = cos(x) + i sin(x)}, even though it was derived from manipulating different series.  That often seems to be a case.  I actually simplified it to sin^2(x) + cos^2(x) = 1 last night by taking the natural log of both sides of the equation to get rid of e and then squaring both sides to get rid of the imaginary part.  The square root ends up just simplifying to that trig identity.

For positive numbers other than one, you can raise them to imaginary powers and get negative numbers.  It is a shocking result.

Edited by Conjurer
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One way to arrive at $$e^{ix}= cos(x)+ i sin(x)$$ is to use the MacLaurin series for $$e^x$$, cos(x), and sin(x).

$$e^x= \sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ \frac{x^2}{2!}+ \frac{x^3}{3!}+ \frax{x^4}{4!}+ \frac{x^5}{5!}+ \cdot\cdot\cdot$$.

$$e^{ix}= 1+ ix+ \frac{(ix)^2}{2!}+ \frac{(ix)^3}{3!}+ \frac{(ix)^4}{4!}+ \frac{(ix)^5}{5!}+ \cdot\cdot\cdot$$.

Of course, $$i^2= -1$$, $$i^3= i^2(i)= -i$$, $$i^4= i^3(i)= -i(i)= 1$$ and then it starts again.

$$= (1- \frac{x^2}{2!}+ \frac{x^3}{3!}- \frac{x^4}{4!}+ \cdot\cdot\cdot)+ i(x- \frac{x^3}{3!}+ \frac{x^5}{5!}+ \cdot\cdot\cdot$$.

And $$cos(x)= 1- \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+ \cdot\cdot\cdot[/tex\] and [tex]sin(x)= x- \frac{x^3}{3!}+ \frac{x^5}{5!}+ \cdot\cdot\cdot$$.

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