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

  • M2★ M3★ M4★ M5

    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

  • M2★ M3★ M4 M5★

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

  • MOD

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

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

    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!)