Calculating $e^{j2\pi t}$ in two ways gives different results: $\cos(2\pi t)+j \sin(2\pi t)$ vs $1^t$

Asked Feb 15 '24 at 09:40

Active Feb 15 '24 at 20:24

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I've encountered the following equations while studying complex exponentials: $$\begin{align} e^{j2\pi t} &= \cos(2\pi t) + j \sin(2\pi t) \tag1\\[4pt] e^{j2\pi t} &= (e^{j2\pi})^t = (\cos(2\pi) + j\sin(2\pi))^t = (1+0)^t = 1^t \tag2 \end{align}$$ Upon comparing (1) and (2), it's evident that they represent different expressions. However, theoretically, these equations should yield the same result.

Could someone help clarify this result?

edited Feb 15 '24 at 09:52

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asked Feb 15 '24 at 09:40

MarounEid

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Why do you think that $e^{j2\pi t}=\left(e^{2j\pi}\right)^t$? Actually, what does $\left(e^{2j\pi}\right)^t$ even mean? – José Carlos Santos Feb 15 '24 at 09:49
This just shows that complex numbers don't behave like real numbers. You cannot apply the rules you learnt for real numbers to the case of complex numbers. – Kavi Rama Murthy Feb 15 '24 at 09:59
1

Hope this solves your problem. https://math.stackexchange.com/a/912686/1277478 – Vinay Karthik Feb 15 '24 at 10:14
Isn't this a duplicate of (many) existing questions here? – GEdgar Feb 15 '24 at 13:21
Related: https://math.stackexchange.com/questions/4488445/how-are-powers-of-complex-numbers-defined/ – Dan Feb 15 '24 at 19:15
Also: https://math.stackexchange.com/questions/136570/determination-of-complex-logarithm – Dan Feb 15 '24 at 20:19

Calculating $e^{j2\pi t}$ in two ways gives different results: $\cos(2\pi t)+j \sin(2\pi t)$ vs $1^t$

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