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

1. Nov 1, 2016

### Prem1998

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?

2. Nov 1, 2016

3. Nov 1, 2016

### robphy

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. Nov 1, 2016

### lurflurf

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. Nov 1, 2016

### Prem1998

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. Nov 1, 2016

### Staff: Mentor

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.