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Let '$i$' be imaginary unit.

$e = e^1$

$e = e^{(2πi/2πi)}$

( because $\frac {2\pi i} {2 \pi i}$ = 1$ )

$e = (e^{(2πi)^{(\frac {1} {2\pi i})}})$

by laws of exponent

$e = (1)^{\frac {1} {2\pi i}}$

( Because $e^{2\pi i} = 1$)

$e = 1$ ( because $1^x = 1$ )

In general , let $Z = r(e)^{ix}$ be any complex number . Using the above result , it can be shown that

$Z = r$

Hence, every complex number is equal to it's modulus? I am in doubt of these two things from some days till I am learning about complex numbers . Please tell what mistake I am doing.

Seeker
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Get_ Maths
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1 Answers1

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This is a classic error. There are similar proofs that $-1=1$. Some exponent laws don’t work (in the same way) in the complex numbers, generally speaking. You need to pay attention to branching issues. In the real numbers there are no problems because the exponential is injective, the logarithm unambiguous. Due to the logarithm’s ill-definedness, $(a^b)^c=a^{bc}$ doesn’t work any more...

Why? Because, if $z=e^w$, then $w=\log z+2\pi in$ for some integer $n$. When you start chaining $z=(e^w)^s=\exp(s\log e^w)$, there is an ambiguity factor of $\exp(s\cdot2\pi i n)$ that must be accounted for. This is what’s causing your problem (but note that, if $s$ is an integer, the ambiguity factor evaluates to one so there’s no problem with integer exponent laws). In more detail, $1^x\equiv1$ isn’t necessarily true, it depends on which branch of the logarithm you use. In your case, we want to say that: $$1^{1/2\pi i}=\exp((1/2\pi i)(\log1+2\pi i n))=\exp(1)=e$$If I choose $n=1$. Crisis averted!

One rule you can always use is $\exp(w+s)=\exp(w)\exp(s)$. That never fails. To be clear, the other rules don’t “fail”, per se, they just need to handled with care. They are correct up to a choice of logarithm.

FShrike
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  • Thanks @Fshrike . Then why we are taught demoivire theorum if that exponent law failed ? – Get_ Maths Aug 13 '22 at 10:17
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    @Get_Maths I’m not sure what you mean. De Moivre’s theorem is perfectly correct! – FShrike Aug 13 '22 at 10:18
  • Integer exponents do work ok, see my correction – FShrike Aug 13 '22 at 10:19
  • We are taught that [e^(ix)]^n = e^(ixn) ; where x is the arguement ; as a proof of demoivire theorum. – Get_ Maths Aug 13 '22 at 10:20
  • It is true for integers. It is ambiguous for non-integer $n$. See my update about how to resolve your particular problem ($1^x\neq1$!) – FShrike Aug 13 '22 at 10:22
  • Thanks . Please can you provide me a proof that it only works for integers ; or is it just a postulate? – Get_ Maths Aug 13 '22 at 10:23
  • @Get_Maths I have. See the part about $(e^w)^s$. We define $a^b$ as $\exp(b\log a)$. Since $\log$ is ambiguous, we get ambiguous answers. That ambiguity is removed if $b$ is an integer! – FShrike Aug 13 '22 at 10:24
  • Thanks , I haven't learnt much yet about logarithms. However, I will revisit it as soon as I'll learn . – Get_ Maths Aug 13 '22 at 10:28
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    @Get_Maths You’re welcome. Good luck! The only thing you need to be aware of, here, is that: I use “branch” to mean “choice of $n$ in $\log z+2\pi i n$” and that: $e^z=e^{z+2\pi i n}$ for any integer $n$ is what is causing problems (not bijective). – FShrike Aug 13 '22 at 10:30