Prove the absolute value of complex exponentials

[pdestud]

I only allowed to prove [MATH] \left| e^{ix} \right| = 1 [/MATH] just by using series apparently. Is it possible to show this proof without refering to [MATH] e^{ix} = \cos(x) + i \sin(x)[/MATH] or the exponential laws [MATH] e^{z+w} [/MATH]? I tried different ways to show it but I think there are errors:

Solution 1:
[MATH] \left| e^{ix} \right| = \sqrt{ \left( \sum_{n = 0}^{\infty} \frac{(ix)^n}{n!} \right) \left( \sum_{n=0}^{\infty} \frac{(-ix)^n}{n!} \right)} [/MATH][MATH] = \sqrt{ \left( 1 + \sum_{n = 1}^{\infty} \frac{(ix)^n}{n!} \right) \left( 1+ \sum_{n = 1}^{\infty} \frac{(-ix)^n}{n!} \right)} [/MATH][MATH] = 1 + \sum_{n = 1}^{\infty} \frac{(-ix)^n}{n!} + \sum_{n = 1}^{\infty} \frac{(ix)^n}{n!} + \sum_{n = 1}^{\infty} \frac{(ix)^n}{n!} \cdot \sum_{n = 1}^{\infty} \frac{(-ix)^n}{n!} [/MATH]
For solution 1, if I expand it, I cannot seem to cancel out terms.

Solution 2:
[MATH] \left| e^{ix} \right| = \sqrt{ \left( \sum_{n = 0}^{\infty} \frac{(ix)^n}{n!} \right) \left( \sum_{n=0}^{\infty} \frac{(-ix)^n}{n!} \right)} [/MATH][MATH] = \sqrt{ \frac{0^0}{0!} + \sum_{n = 1}^{\infty} \frac{0^n}{n!} } [/MATH]
For solution 2, the first term is undefined.

Solution 3:
[MATH] \left| e^{ix} \right| = \sqrt{ \left( \sum_{n = 0}^{\infty} \frac{(ix)^n}{n!} \right) \left( \sum_{n=0}^{\infty} \frac{(-ix)^n}{n!} \right)} [/MATH][MATH] = \sqrt{ \sum_{n = 0}^{\infty} \frac{0^n}{n!} } [/MATH][MATH] = e^{0} [/MATH][MATH] = 1 [/MATH]
For solution 3, the sum doesn't make sense because I should be summing zeroes. Wouldn't it mean my infinite sum is also zero? Plus, from solution 2, for n = 0, it should be undefined.

Are there any other possibilities? Please clarify and thanks.

 

[pka]

I only allowed to prove [MATH] \left| e^{ix} \right| = 1 [/MATH] just by using series apparently. Is it possible to show this proof without refering to [MATH] e^{ix} = \cos(x) + i \sin(x)[/MATH] or the exponential laws [MATH] e^{z+w} [/MATH]? I tried different ways to show it but I think there are errors:

This is one persons opinion but strongly held: it is nonsense not to use \(\displaystyle e^{ix} = \cos(x) + i \sin(x)\)
Years & years of advanced calculus classes have had the pleasure of seeing that using series:
\(\displaystyle \sin (x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{(2n + 1)!}}} ,\quad \cos (x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}},\quad \exp (x) = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{(n)!}}} } \)
we arrive at that such a beautifully simple expression.
So use the given series to do this question.