# Understanding Euler's Identity

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?

• $\exp(a) = \exp(b)$ 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. – user127.0.0.1 Jan 6 '14 at 19:29
• Why do you use $j$ for $i$? – TMM Jan 6 '14 at 19:30
• @TMM it's quite usual in electrical engineering fo example – user127.0.0.1 Jan 6 '14 at 19:31
• 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. – copper.hat Jan 6 '14 at 19:43
• Exponential functions of a REAL number are one-to-one. Exponential functinos of a complex number are not. – Michael Hardy Jan 6 '14 at 20:22

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 $$\exp(z) = \underbrace{\exp(\Re z)}_{\exp: \mathbb R\to\mathbb R} \cdot (\cos(\Im z) + i\sin(\Im z))$$
• However you may call it: $$\exp(z) = \exp(\Re z) (\cos(\Im z) + i\sin(\Im z))$$ – AlexR Jan 6 '14 at 19:32
• @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. – Erick Wong Jan 6 '14 at 19:34
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 $2\pi\ne0$?