# Does iωLog[(-iω + √((r^2-ω^2)/r)] = ArcTan[ω/√(r^2-ω^2)]?

1. Jun 19, 2014

### Katy_A

This is my first post so I apologise for the notation not being very good.

I have been given some notes to check over by my PhD supervisor, and the notes he has given me state that:

$$i \omega \log \left(\frac{\sqrt{r^2-\omega ^2}-i \omega }{r}\right)$$ = $$\tan ^{-1}\left(\frac{\omega }{\sqrt{r^2-\omega ^2}}\right)$$

But I can't seem to reproduce this answer using ArcTan(x) = (i/2)ln(1-ix/1+ix). Can someone please either show me how to obtain this ArcTan result from the exponential equation I have given or tell me that it's wrong.

Thank you.

Last edited: Jun 19, 2014
2. Jun 19, 2014

### Staff: Mentor

so x = w/sqrt(r^2 -w^2) from arctan(x) side

try subbing the x expression above into the righthand side of the answer and see if you can reduce it to an expression in w that matches the initial log expression from the first equation. That would demonstrate that the answer is correct.

You could also use matlab to compare the two expressions plotting both against x to see if the curves look the same as a kind visual proof that the answer is true before you begin the algebraic work trying to prove it.

3. Jun 19, 2014

### Katy_A

I am using Mathematica and I have plotted the graphs against one another and they do appear to be the same. But I still can't work out why by using pencil and paper by substituting in for x.

And Mathematica gives $$\tan ^{-1}\left(\frac{\omega }{\sqrt{r^2-\omega ^2}}\right)$$ = $$\frac{1}{2} i \log \left(1-\frac{i \omega }{\sqrt{r^2-\omega ^2}}\right)-\frac{1}{2} i \log \left(1+\frac{i \omega }{\sqrt{r^2-\omega ^2}}\right)$$...so I don't understand why the graphs says they are equal.

4. Jun 19, 2014

### Staff: Mentor

I don't have the thing solved either so I'm looking around for the key step:

Have you looked at Euler's formula?

http://en.wikipedia.org/wiki/Euler_formula

Roger Cote's formula is shown which mixes log and trig functions together:

ln(cos x + isin x)=ix

Last edited: Jun 19, 2014
5. Jun 19, 2014

### Saitama

Using the formula you posted, you get:
$$\tan^{-1}\left(\frac{\omega}{\sqrt{r^2-x^2}}\right)=\frac{i}{2}\ln\left(\frac{\sqrt{r^2-x^2}-i\omega}{\sqrt{r^2-x^2}+i\omega}\right)$$
Inside the log, multiply and divide by $\sqrt{r^2-\omega^2}-i\omega$.
$$\frac{i}{2}\ln\left(\frac{\left(\sqrt{r^2-x^2}-i\omega\right)^2}{r^2-\omega^2+\omega^2}\right)=\frac{i}{2}\ln\left(\frac{\sqrt{r^2-\omega^2}-i\omega}{r}\right)^2=i\ln\left(\frac{\sqrt{r^2-\omega^2}-i\omega}{r}\right)$$

But this doesn't give the extra factor of $\omega$ in OP.

6. Jun 19, 2014

### Bill Simpson

I must be missing some key bit here

Code (Text):
In[1]:= r = RandomReal[{-10, 10}]

Out[1]= -6.19522

In[2]:= w = RandomReal[{-10, 10}]

Out[2]= -4.22769

In[3]:= I w Log[(Sqrt[r^2 - w^2] - I w)/r]

Out[3]= -10.1064 - 9.38736*10^-16 I

In[4]:= ArcTan[w/Sqrt[r^2 - w^2]]

Out[4]= -0.751057
If you show exactly what you did to plot these then perhaps the problem can be discovered.

7. Jun 29, 2014

### Newton55

I'm with Pranav-Arora, i can't get the ω.. Maybe the plot isn't correct at all..