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Euler's formula is

$$ e^{i \theta } = \cos \theta + i \sin \theta $$

If we take $\theta$ to be $2 \pi$ we get

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

Which simplifies to

$$e^{2i \pi} = 1$$

But we also know

$$e^0 = 1$$

Therefore, can we not say

$$e^0 = e^{2i \pi}$$

and thus,

$$0 = 2i \pi$$

Which leads to being a big problem for all imaginary numbers.

Example with $25i$:

$$\begin{align} 25i &= \frac{25 (2 \pi) i}{2 \pi} \\ 25i &= \frac{25 * 0}{2 \pi} \\ 25i &= 0 \\ \end{align}$$

MaximusFastidiousIrreverence
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    If $a,,b$ are real numbers then $e^a=e^b$ if and only if $a=b$, however, this is not true if $a,,b$ are complex numbers. – John Wayland Bales Mar 25 '18 at 04:30
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    What if instead of exponentials, we just considered some arbitrary function. If $f(a)=f(b)$, can we say that $a=b$? No, because there may be multiple values that map to $f(a)$ (such as $a,b,...$). – QC_QAOA Mar 25 '18 at 04:31
  • First displayed equation is incorrect. Surely you mean $i\theta$ instead of $1\theta$. – MPW Mar 25 '18 at 04:53

2 Answers2

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You cannot say $e^{0}=e^{2\pi i}$ implies $0=2\pi i$.

The map $z\mapsto e^{z}$ is not one-to-one in the complex numbers.

parsiad
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  • Fun fact, an easter egg from the God: it holds that $0 = 1 + e^{i\pi}$. All magical numbers in one equation: 0, 1, e, i, $\pi$ – Aelx Dec 07 '20 at 22:53
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The correct conclusion of $$e^a=e^b$$ is not that $a=b$, but rather that $a=b+2k\pi i$ for some $k\in\mathbb Z$.

MPW
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