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

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      modestwallaby M2★ M3★ M4★ M5 last edited by

      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

      E quacker88 2 Replies Last reply Reply Quote 5
      • E
        energizedpanda M2★ M3★ M4 M5★ @modestwallaby last edited by energizedpanda

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

        ~∑nergized Pand∆

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        • quacker88
          quacker88 MOD @modestwallaby last edited by

          @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 1 Reply Last reply Reply Quote 3
          • quacker88
            quacker88 MOD @quacker88 last edited by

            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!)

            1 Reply Last reply Reply Quote 2

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