Math also Fails! -2

Most of you shall be knowing about Euler's formula: $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ If you put $\theta=\tau$ (where $\tau$ is a unit of measuring angle if you don't know about it see this and this) $e^{i\tau}=\cos(\tau)+i\sin(\tau)=1+i\times0=1$ $\Rightarrow \boxed{e^{i\tau}=1}$ Taking $\ln$ on both sides $\ln(e^{i\tau})=\ln(1)$ $\Rightarrow i\tau\ln(e)=\ln(1)$ $\Rightarrow i\tau\times1=\ln(1)$ As $\ln(1)=0$ $\therefore i\tau=0$ As $i≠0$ $\therefore {\red{\tau=0}}$ If you want you can further write $2\pi=0$ $\Rightarrow \red{\pi=0}$ Now will you agree with the last statement? Doesn't this means Math also fails!

Note:

$i$ denotes the imaginary unit
See Math Also Fails! -1

#Algebra

Easy Math Editor

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Comments

The flaw is that in the complex plane, $e^{z_1} = e^{z_2} \not \Rightarrow z_1 = z_2$ . This is because $e^{i\theta} = \cos \theta + i \sin \theta$ . The sine and cosine functions each have a period of $2\pi$ , so $e^{i\theta} = e^{i(\theta + 2\pi)}$ . This leads to the misleading conclusion that $0 = 2\pi$ .

Elijah L - 1 year ago

Now, that's the correct explanation!

Zakir Husain - 1 year ago

Wow!

Mahdi Raza - 1 year ago

@Elijah L although an argunent of 2π is not allowed in argand plane ,but we can generalize by considering periodicity.