# 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 Note by Zakir Husain
3 months, 2 weeks ago

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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 $0 = 2\pi$.

- 3 months, 2 weeks ago

Now, that's the correct explanation!

- 3 months, 2 weeks ago

Wow!

- 3 months, 2 weeks ago

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

- 3 months ago

@Zakir Husain The problem starts when you forget about periodicity of trignometric functions and limits of argument of complex number.

we know that, e^{iθ}=cos(θ)+isin(θ) .The argument of this complex number will be θ and the argument of a complex number should be in the range of (-π,π].So,you can't take θ as 2π,-π etc .

Now,if we want to expand its range we should consider periodicity of trignometric function. e^{iθ}=e^{i2nπ+θ}

I know this question is not for me and neither i have mentioned to answer it.But,as a noob i have tried to figure out something.Please comment if you find any error.

- 3 months ago

- 3 months ago

It is also an explanation same as that given by @Elijah L, but stated differently.

- 3 months ago

Oh,i see but i am still confused that argument could be 2π or not.

- 3 months ago

It can be, but you need to take this rule with you :

If $\boxed{\blue{e^{z_1}=e^{z_2}}:z_1\in C, z_2\in C}\cancel{\Rightarrow}\boxed{\red{z_1=z_2}}$

- 3 months ago

I know it .But,is it a rule?? Many books say that argument has limit(-π,π]

- 3 months ago

- 3 months, 2 weeks ago

@Brilliant Mathematics, I just encountered a bug where i did not get a notification for this mention.

- 3 months, 2 weeks ago

@Zakir Husain, there is a small little typo in the second line as well. There should be a closed bracket instead of 0

- 3 months, 2 weeks ago

@Zakir Husain, Is there a part 1 or 3 of "Math Also Fails"

- 3 months, 2 weeks ago

@Mahdi Raza- Math Also Fails! -1, No part 3 is posted yet

- 3 months, 2 weeks ago

Ok

- 3 months, 2 weeks ago

I don't understand much of the proof, but in my opinion there has to be some problem in using $\sqrt{-1}$ as I saw some video on Numberphile which says that some properties of numbers are not allowed for $\sqrt{-1}$.

- 3 months, 2 weeks ago

@Vinayak Srivastava - I onlly used $2$ properties:

• The euler's identity

• If $ai=0$ then $a=0$ because $i≠0$

These are all applicable for $i$.

- 3 months, 2 weeks ago

@Zakir Husain, I don't agree with the last statement at all. Also (not being rude, this is a good note, don't take it personally), I believe that the Math also Fails series is somewhat undermining mathematics in a way inexpressible. For example, if you're saying $\pi = 0$, you're in effect, undermining international $\pi$ day and all the people who have worked hard to help improve the value of $\pi$. So, I implore you, although this is a joke and I believe this is a good one, don't put a post that undermines people's work by, in effect, proving mathematics wrong (I check all new discussions every day - that's why I'm saying not to post a note that undermines people's work by, in effect, proving mathematics wrong). You can still post it, as long as it steers away from people's work (i.e. don't do $e$, number sequences that people found etc.) and my defensive opinion of people's hard work and mathematics and it's reputation...

- 3 months, 2 weeks ago

@Yajat Shamji - Mathematics demands proof. I had just used legal mathematical arguments, if you can find a flaw then find it!

- 3 months, 2 weeks ago

Flaw on $4$th step...

- 3 months, 2 weeks ago

Please write the expression so I will be able to understand

- 3 months, 2 weeks ago

$ln(e^{ir}) = ln(1) \rightarrow irln(e) = ln(1)$ is flawed.

- 3 months, 2 weeks ago

It is a legal mathematical property of logarithm see here for more details

- 3 months, 2 weeks ago

Yo I think you're taking this too seriously. Of course it's not seriously questioning the statement that $\pi = 3.14$.

- 3 months, 2 weeks ago

The way I see it, it's not a problem at all. We all know that the proof is false, and it's a sort of entertaining mathematical challenge to find out where the error lies.

- 3 months, 2 weeks ago

Yeah exactly, the guy above can't seem to get it.

- 3 months, 2 weeks ago