Need help understanding Euler's formula (complex numbers)

[dmehling]

I have a couple questions regarding Euler's formula. First I'm confused about the notation e^i\vartheta. To me the notation implies that we are raising e to the exponent i multiplied by \vartheta. Is this correct? If so, how would you do that? Also, my second question regards the second part of that equation: cos\vartheta + isin\vartheta. That makes sense to me, but sometimes I see it written as cosx + isinx. I don't understand what that means. What would be the cosine or sine of x?

 

[mathman]

You seemed to be confused about mathematical notation. e^ix= cosx+isinx is a relationship where x is a dummy symbol. You can plug in any value (make sure you understand the trig function arguments are radians, not degrees) on both sides and the relationship holds. It doesn't matter what symbol you use as long as it is the same for all three functions.

 

[dmehling]

Thanks for answering my second question. I didn't realize that variables could be used to stand in for something like an angle. I figured that simply using theta would suffice as a variable of sorts.

What about my first question? Is the first part of the formula an exponential expression or not, or is this an expression that cannot be evaluated?

 

[elibj123]

You can't really define the exponent of a complex number.
This formula can be derived by several tools, but the use of these tool in the field of complex number is not rigorous: you can't really prove that they behave the same in the complex field as they in the real field (just like the exponent itself), so this formula remains more like a definition rather than a theorem.

 

[HallsofIvy]

elibj123, everything in mathematics is "rigorous". We can and do "really define the exponent of a complex number".

You can define exponentials or things other than regular real numbers, not just imaginary numbers and complex numbers, but matrices, etc. by using the Taylor's series.

It is shown in calculus that the power series
\sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ \frac{1}{2}x^2+ \frac{1}{3!}x^3+ \cdot\cdot\cdot[/itex] <br /> converges to e^x for all x.<br /> <br /> Since that involves only products and sums, we can use that to define e^A for anything we can define products and sums for (including complex numbers and matrices, etc.).<br /> <br /> In particular, we know that (ix)^2= i^2x^2= -x, (ix)^3= (ix)^2(ix)= (-x)(ix)= -ix^3, and (ix)^4= (ix)^3(ix)= (-ix^3)(ix)= -(i^2)x^4= x^4. Of course, higher powers just repeat that. In particular:<br /> e^{ix}= 1+ (ix)+ \frac{1}{2}(ix)^2+ \frac{1}{3!}(ix)^3+ \frac{1}{4!}x^4+ \frac{1}{5!}(ix)^5 \cdot\cdot\cdot<br /> = 1+ ix- \frac{1}{2}(-x^2)+ \frac{1}{3!}(-ix^3)+ \frac{1}{4}(x^4}+ \frac{1}{5!}ix^5+ \cdot\cdot\cdot<br /> <br /> We can divide that into &quot;real&quot; and &quot;imaginary&quot; parts:<br /> e^{ix}= (1- \frac{1}{2}x^2+ \frac{1}{4!}x^4-\cdot\cdot\cdot)+ i(x- \frac{1}{3!}x^4+ \frac{1}{5!}x^5+\cdot\cdot\cdot<br /> and recognize those as the Taylor&#039;s series for cosine and sine:<br /> e^{ix}= cos(x)+ i sin(x).