2
$\begingroup$

I would like to understand one specific moment in Euler's Identity, namely

$$e^{j\theta}=\cos(\theta)+j\sin(\theta)$$

where $j=\sqrt{-1}$. We also know that

$$e^{j2(\pi)}=\cos(2\pi)+j\sin(2\pi)$$

but $\sin(2\pi)=0$ and $\cos(2\pi)=1$, and $1=e^{0}$, so we get that $e^{j2(\pi)}=e^{0}$.

But we get that $j2\pi=0$ which means that $j=0$, but on the other hand $j=\sqrt{-1}$. I want to ask one question: why is it allowed to use such symbols in identity, which finally may cause some strange equality?

$\endgroup$
  • 1
    $\begingroup$ $\exp(a) = \exp(b)$ does not imply $a=b$. Check your definition of $\exp(a)$ and note, that potential rules (is this the correct english word for it?) are not valid for complex numbers in general. $\endgroup$ – user127.0.0.1 Jan 6 '14 at 19:29
  • $\begingroup$ Why do you use $j$ for $i$? $\endgroup$ – TMM Jan 6 '14 at 19:30
  • $\begingroup$ @TMM it's quite usual in electrical engineering fo example $\endgroup$ – user127.0.0.1 Jan 6 '14 at 19:31
  • $\begingroup$ It is used in electrical engineering, although since $j$ is also used (current densities), it is not clear to me why the switch was made. $\endgroup$ – copper.hat Jan 6 '14 at 19:43
  • $\begingroup$ Exponential functions of a REAL number are one-to-one. Exponential functinos of a complex number are not. $\endgroup$ – Michael Hardy Jan 6 '14 at 20:22
6
$\begingroup$

The mistake is that $\exp: \mathbb C \to \mathbb C$ is no longer a bijection. Thus $$e^{2\pi i} = e^{0} \not\Rightarrow 2\pi i = 0$$ In general $$\exp(z) = \exp(z+2\pi i) \qquad \forall\ z\in\mathbb C$$ because of the periodicity of $\sin$ and $\cos$ and the definition $$\exp(z) = \underbrace{\exp(\Re z)}_{\exp: \mathbb R\to\mathbb R} \cdot (\cos(\Im z) + i\sin(\Im z))$$

$\endgroup$
  • $\begingroup$ because of rotation principle right? $\endgroup$ – dato datuashvili Jan 6 '14 at 19:29
  • $\begingroup$ However you may call it: $$\exp(z) = \exp(\Re z) (\cos(\Im z) + i\sin(\Im z))$$ $\endgroup$ – AlexR Jan 6 '14 at 19:32
  • $\begingroup$ thanks very much for answer,happy new year $\endgroup$ – dato datuashvili Jan 6 '14 at 19:34
  • $\begingroup$ @datodatuashvili Perhaps, but I would sooner say it's just unwarranted to assume that $e^x = e^y$ means $x=y$ for complex numbers just because you're used to that being true for real numbers. $\endgroup$ – Erick Wong Jan 6 '14 at 19:34
  • 1
    $\begingroup$ @AlexR it was marked as answered $\endgroup$ – dato datuashvili Jan 6 '14 at 19:45
2
$\begingroup$

You are worried that $e^{j\cdot 2\pi}=\cos(2\pi)+j\sin(2\pi)=\cos(0)+j\sin(0)=e^0$, even though $j\cdot2\pi\ne0$. Does it also worry you that $$\cos(2\pi)=\cos(0)$$ and $$\sin(2\pi)=\sin(0),$$ even though $2\pi\ne0$?

$\endgroup$
  • 1
    $\begingroup$ yes yes right,just i was thinking about this whole day and now i am relaxed :D $\endgroup$ – dato datuashvili Jan 6 '14 at 19:36
  • $\begingroup$ Ok, very good :) $\endgroup$ – Carsten S Jan 6 '14 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.