Either ( a^0=1) is wrong or (e^ix=cos x +i sin x) is wrong

[modestwallaby]

If a^0=1, and e^ix=cos x +i sin x, then
e^i2π=cos 2π+i sin 2π=1+0=1

But 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

[energizedpanda]

@modestwallaby very interesting... I've never thought about this before

[quacker88]

@modestwallaby I'm not an expert on this subject, but I believe this is similar logic to the following:

$$\sin(2\pi)=\sin(0)=0$$ $$2\pi=0$$

[quacker88]

I did some searching around and here's what I concluded:

For some functions, if

$$f(x)=f(y), \text{this does NOT mean that } x=y$$

.
For example, look at

$$f(x)=x^2$$ $$f(-1)=f(1) \text{ , however} -1 \neq 1$$

Same 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!)