# Argument of a complex expression

Homework Statement:
What is the correct way of computing the argument of the following equation?
Relevant Equations:
I am trying to compute the argument ##\Phi## of the equation

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

which using Euler's equation can also be written in the form

$$\exp\left[i\left(\pi+\varphi\right)\right]\frac{\tau-r\exp\left(-i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)} \tag{2}$$
Problem Statement: What is the correct way of computing the argument of the following equation?
Relevant Equations: I am trying to compute the argument ##\Phi## of the equation

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

which using Euler's equation can also be written in the form

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

(1) If we use the first equation, we can first separate out the real and imaginary parts of the expression by multiplying by the complex conjugate of the denominator

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

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

Since, for a complex number ##z##, ##\text{arg}\left(z\right)=\text{atan }\left[\Im\left(z\right)/\Re\left(z\right)\right]##, we have:

$$\Phi=\text{atan}\left[\frac{-\tau\left(1-r^{2}\right)\sin\varphi}{r+\tau^{2}r-\tau\left(r^{2}+1\right)\cos\varphi}\right].$$

(2) However, the paper I am looking at used the second form (equation (2)), which readily gives:

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

Clearly, these two answers are very different. Which method is correct, and what is the cause of the discrepancy? Shouldn't we end up with the same expression for the argument regardless of the form we start with?

Any explanation is appreciated.

Mark44
Mentor
Problem Statement: What is the correct way of computing the argument of the following equation?
Relevant Equations: I am trying to compute the argument ##\Phi## of the equation

$$\frac{r-\tau\exp\left(i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)} \tag{1}$$
Minor point: this is not an equation.
roam said:
which using Euler's equation can also be written in the form

$$\exp\left[i\left(\pi+\varphi\right)\right]\frac{\tau-r\exp\left(-i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)} \tag{2}$$
I would do something different -- replace ##\exp(i\phi)## in expression (1) by ##\cos(\phi) + i\sin(\phi)##, and then multiply by the complex conjugate of the denominator ##(1 + \tau r \exp(i\phi))## over itself.
I haven't worked this through, but that's where I would start.
roam said:
(1) If we use the first equation, we can first separate out the real and imaginary parts of the expression by multiplying by the complex conjugate of the denominator

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

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

Since, for a complex number ##z##, ##\text{arg}\left(z\right)=\text{atan }\left[\Im\left(z\right)/\Re\left(z\right)\right]##, we have:

$$\Phi=\text{atan}\left[\frac{-\tau\left(1-r^{2}\right)\sin\varphi}{r+\tau^{2}r-\tau\left(r^{2}+1\right)\cos\varphi}\right].$$

(2) However, the paper I am looking at used the second form (equation (2)), which readily gives:

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

Clearly, these two answers are very different. Which method is correct, and what is the cause of the discrepancy? Shouldn't we end up with the same expression for the argument regardless of the form we start with?

Any explanation is appreciated.

Minor point: this is not an equation.
I would do something different -- replace ##\exp(i\phi)## in expression (1) by ##\cos(\phi) + i\sin(\phi)##, and then multiply by the complex conjugate of the denominator ##(1 + \tau r \exp(i\phi))## over itself.
I haven't worked this through, but that's where I would start.

Hi @Mark44

$$z:=\frac{r-\tau\exp\left(i\varphi\right)}{1-\tau r\exp\left(i\varphi\right)}=\frac{r-\tau\left(\cos\varphi+i\sin\varphi\right)}{1-\tau r\exp\left(i\varphi\right)}$$

Multiplying with the complex conjugate:

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

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

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

which gives:

##\Re\left(z\right)=\frac{r+\tau r^{2}\cos\varphi-\tau-\tau^{2}r\cos\varphi+\tau^{2}r\sin^{2}\varphi}{1-\tau^{2}r^{2}\exp\left(2i\varphi\right)}##
##\Im\left(z\right)=\frac{\tau r^{2}\sin\varphi-i\tau r\sin\varphi\cos\varphi}{1-\tau^{2}r^{2}\exp\left(2i\varphi\right)}##

Therefore,

$$\text{arg}\left(z\right)=\text{atan}\left(\frac{\tau r^{2}\sin\varphi-\tau^{2}r\sin\varphi\cos\varphi}{r+\tau r^{2}\cos\varphi-\tau-\tau^{2}r\cos\varphi+\tau^{2}r\sin^{2}\varphi}\right).$$

Is this reducible to the form for ##\Phi## given in my post #1? I am not sure how it can be manipulated further.

Gaussian97
Homework Helper
$$=\frac{r+\tau^{2}r-\tau\left(r^{2}+1\right)\cos\varphi}{1-\tau r\cos\varphi+\left(\tau r\right)^{2}}+i\frac{-\tau\left(1-r^{2}\right)\sin\varphi}{1-\tau r\cos\varphi+\left(\tau r\right)^{2}}.$$
I think this is correct except for a 2 in the denominator that doesn't affect the result.