What's wrong with this proof of sin(i)=0?

  • B
  • Thread starter Prem1998
  • Start date
  • Tags
    Proof
In summary, The proof of sin(i)=0 is considered incorrect because it violates the fundamental rules of complex numbers. It cannot be fixed or corrected and using it can lead to false conclusions and errors. The correct approach to proving sin(i)=0 is to use the Taylor series expansion of the sine function, which converges for all complex numbers.
  • #1
Prem1998
148
13
We have,
e^(ix)=cosx+isinx
So, e^(i*i)=cosi+isini
Or e^-1=cosi+isini
Or 1/e + 0*i= cosi+isini
So, cosi=1/e and sini=0
But that's not the value of sin(i) that I found on the internet. These values are not even satisfying cos^2(x)+sin^2(x)=1.
What did I miss?
 
Mathematics news on Phys.org
  • #3
Prem1998 said:
We have,
e^(ix)=cosx+isinx
So, e^(i*i)=cosi+isini
Or e^-1=cosi+isini
Or 1/e + 0*i= cosi+isini
So, cosi=1/e and sini=0
But that's not the value of sin(i) that I found on the internet. These values are not even satisfying cos^2(x)+sin^2(x)=1.
What did I miss?
e^-1=cosi+isini
1/e + 0*i= cosi+isini

On the left, you have the real and imaginary parts of ##e^{-1}##.
So, you should find the real and imaginary parts on the right-hand side.
Since ##\cos(i)=\cosh 1## and ##\sin(i)=i\sinh(1)##
http://www.wolframalpha.com/input/?i=cos(i)
http://www.wolframalpha.com/input/?i=sin(i)
we have:
##\begin{align*}
1/e + 0i
&= \cosh 1+i (i\sinh 1)\\
&= \cosh 1-\sinh 1\\
&= \frac{e^1+e^{-1}}{2}-\frac{e^1-e^{-1}}{2}\\
&= e^{-1}
\end{align*}##
 
  • #4
This is ok

We have,
e^(ix)=cosx+isinx
So, e^(i*i)=cosi+isini
Or e^-1=cosi+isini
Or 1/e + 0*i= cosi+isini

This is false
So, cosi=1/e and sini=0

in fact
cosi=cosh1
isini=sinh1
which are both real

But that's not the value of sin(i) that I found on the internet. These values are not even satisfying cos^2(x)+sin^2(x)=1.
What did I miss?

what value did you find? The correct values I gave above satisfy cos^2(x)+sin^2(x)=1.
 
  • #5
lurflurf said:
This is ok

We have,
e^(ix)=cosx+isinx
So, e^(i*i)=cosi+isini
Or e^-1=cosi+isini
Or 1/e + 0*i= cosi+isini

This is false
So, cosi=1/e and sini=0

in fact
cosi=cosh1
isini=sinh1
which are both real

But that's not the value of sin(i) that I found on the internet. These values are not even satisfying cos^2(x)+sin^2(x)=1.
What did I miss?

what value did you find? The correct values I gave above satisfy cos^2(x)+sin^2(x)=1.
So, the problem was when I compared both sides considering cosi and sini to be real, right?
One more thing, does the sine of complex numbers have any physical meaning? Just like how the sine of real numbers represents the y-co-ordinate of the point that we reach on a unit circle by rotating throught that angle. Do they mean anything or are they just useless numbers obtained by plugging in complex numbers in the tailer series?
 
  • #6
Prem1998 said:
So, the problem was when I compared both sides considering cosi and sini to be real, right?
Right. Cosine and sine are complex in general - only for real arguments they are real.

I'm not aware of physical applications of complex arguments for the sine. Damped oscillations can be described with a complex period in the exponential, but converting that to a sine and cosine does not really help.
 

Related to What's wrong with this proof of sin(i)=0?

1. What is the proof of sin(i)=0?

The proof of sin(i)=0 is a mathematical calculation that attempts to show that the sine of the imaginary number i is equal to 0.

2. Why is this proof considered incorrect?

This proof is considered incorrect because it violates the fundamental rules of complex numbers. In particular, it involves taking the logarithm of a negative number, which is undefined in the complex plane.

3. Can this proof be fixed or corrected?

No, this proof cannot be fixed or corrected. The mistake is inherent in the fundamental rules of complex numbers and cannot be resolved without completely redefining those rules.

4. What are the consequences of using this incorrect proof?

Using this incorrect proof can lead to false conclusions and errors in further calculations. It also undermines the integrity of mathematical principles and can cause confusion among students and researchers.

5. What is the correct approach to proving sin(i)=0?

The correct approach to proving sin(i)=0 is to use the Taylor series expansion of the sine function, which converges for all complex numbers. This approach takes into account the unique properties of complex numbers and avoids the pitfalls of the incorrect proof.

Similar threads

  • General Math
Replies
4
Views
2K
Replies
8
Views
3K
  • General Math
Replies
1
Views
3K
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
857
  • General Math
Replies
7
Views
1K
  • General Math
2
Replies
66
Views
4K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • General Math
Replies
11
Views
1K
Back
Top