I am trying to verify a relationship presented in this paper. I want to verify that the argument of

$$\frac{E_{2}}{E_{1}}=\frac{r-\tau\exp\left(i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)}, \tag{1}$$

is equal to

$$\Phi=arg\left(\frac{E_{2}}{E_{1}}\right)=\pi+\varphi+\text{atan}\left(\frac{r\sin\varphi}{\tau-r\cos\varphi}\right)+\text{atan}\left(\frac{r\tau\sin\varphi}{1-r\tau\cos\varphi}\right). \tag{2}$$


We know that

$$\Phi=arg\left(\frac{E_{2}}{E_{1}}\right)=\text{atan}\left[\frac{\text{Im}\left(\frac{E_{2}}{E_{1}}\right)}{\text{Re}\left(\frac{E_{2}}{E_{1}}\right)}\right]. \tag{3}$$

So, the first step would be to separate out the real and imaginary parts of (1) by multiplying by the complex conjugate of the denominator:

$$\frac{r-\tau\exp\left(i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)}.\frac{1-\tau r\exp\left(-i\varphi\right)}{1-\tau r\exp\left(-i\varphi\right)}=\frac{r-\tau r^{2}\exp\left(-i\varphi\right)-\tau\exp\left(i\varphi\right)+\tau^{2}r}{1-\tau r\left[\exp\left(i\varphi\right)+\exp\left(-i\varphi\right)\right]+\left(\tau r\right)^{2}}$$

$$=\frac{r+\tau^{2}r-\tau\left(r^{2}+1\right)\cos\varphi}{1-\tau r\cos\varphi+\left(\tau r\right)^{2}}+i\frac{-\tau\left(1-r^{2}\right)\sin\varphi}{1-\tau r\cos\varphi+\left(\tau r\right)^{2}}.$$

Now using (3) we find the argument to be:


However, this is clearly different from the desired expression given in (2).

So, is this a mistake on my part or on the part of the authors? Is it possible to arrive at equation (2)?

Any explanation would be greatly appreciated.

P.S. The authors have used Euler's equation to express (1) as $\frac{E_{2}}{E_{1}}=\exp\left[i\left(\pi+\varphi\right)\right]\frac{\tau-r\exp\left(-i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)}$.

  • $\begingroup$ Try using $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$ $\endgroup$
    – saulspatz
    May 23 '19 at 13:53

Using the expression in equation $(4)$ of the paper, $$\frac{E_{2}}{E_{1}}=\exp\left[i\left(\pi+\varphi\right)\right]\frac{\tau-r\exp\left(-i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)}\tag{4}$$ and applying $$\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$$ we get $$\arg\left({E_1\over E_2}\right)=\pi+\varphi+\arg(\tau-re^{-i\theta})-\arg(1-r\tau e^{i\theta})$$

Now computing the args by separating out the real and imaginary parts in the last two terms gives the desired formula.

  • $\begingroup$ Thanks a lot, that works perfectly. I do have a follow-up question: my equation (1) is essentially identical to equation (4) of the paper. By using $e^{i\pi}=-1$ we can transform one equation to the other. So, what might be the cause of the discrepancy between this result and my result above? Shouldn't we obtain the same result regardless of the form we start with? $\endgroup$
    – Nick
    May 24 '19 at 7:27
  • $\begingroup$ math.stackexchange.com/questions/474992/… $\endgroup$
    – saulspatz
    May 24 '19 at 12:40

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