Either ( a^0=1) is wrong or (e^ix=cos x +i sin x) is wrong
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If a^0=1, and e^ix=cos x +i sin x, then
e^i2π=cos 2π+i sin 2π=1+0=1But because e^0=1, so i2π=0, so either i =0, or π=0, which we know is impossible, so either a^0=1 is wrong, or e^ix=cos x+ i sin x is wrong
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@modestwallaby very interesting... I've never thought about this before
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@modestwallaby I'm not an expert on this subject, but I believe this is similar logic to the following:
$$\sin(2\pi)=\sin(0)=0 $$ $$2\pi=0 $$ -
I did some searching around and here's what I concluded:
For some functions, if
$$f(x)=f(y), \text{this does NOT mean that } x=y $$.
$$f(x)=x^2 $$ $$f(-1)=f(1) \text{ , however} -1 \neq 1 $$
For example, look atSame follows for sin(x)
$$\sin(0)=\sin(2\pi)=\sin(4\pi)=\sin(1290\pi)=0 $$and obviously the values inside are not equal to each other.
This is the case with
$$f(x)=e^{ix} $$You cannot deduce that just because
$$e^{ 2\pi \cdot i} = e^{0\cdot i} \implies 2\pi=0 $$This is almost like saying
$$1^{100} = 1^1 \implies 100=1 $$Exponentiation simply just doesn't work that way. I hope this explanation helps you! Let me know if you have any more questions (this is pretty confusing, so don't worry!)
