1
$\begingroup$

I am having some troubles in understanding how to solve (if possible) this complex equation:

$$A a e^{-aT} = \frac{2i}{3\Re(T)}\left(\omega_0 + A e^{-aT}\right)$$

where

  • $A, a, \omega_0$ are real constants;

  • $i$ is the imaginary unit;

  • $T$ is a complex number, and it can be willingly written as $T = \sigma + i\zeta$;

  • $\Re(T)$ is the real part of $T$.

I tried to write $\Re(T)$ as $\sigma$, or also as $\frac{T + T^*}{2}$ but this did not bring me to a solution.

Is there some way to solve this equation for $T$?

I also tried to use a Taylor series for the exponential but I actually don't know what the constants are (they may be small quantities or big ones or some and some), hence it is not a good way to proceed.

Thanks in advance!

  • 0
    You want to solve for $T$? – 2017-01-18
  • 0
    @SimpleArt Yesss! – 2017-01-18
  • 0
    :-( Tried Lambert W function, but no matter what, the $\Re(T)$ is being a pain. – 2017-01-18
  • 0
    @SimpleArt Indeed, is that term that makes everything strange! – 2017-01-18
  • 0
    have you tried to seperate real and imaginary part? – 2017-01-18
  • 0
    @tired Separate how? They are the argument of an exponential. At most I can write $T = \sigma + i\zeta$ as written, but the exponential term stays. – 2017-01-18

1 Answers 1

2

An idea is to begin seperating the real and imaginary parts (this was too long to put in a comment). Use:

$$\exp\left[-\text{a}\cdot\left(\sigma+\zeta i\right)\right]=\exp\left[-\text{a}\sigma\right]\cdot\exp\left[-\text{a}\zeta i\right]=\frac{\cos\left(\text{a}\zeta\right)-\sin\left(\text{a}\zeta\right)i}{\exp\left[\text{a}\sigma\right]}\tag1$$

So, we get for the LHS:

1.$$\Re\left(\text{A}\cdot\text{a}\cdot e^{-\text{a}\cdot\text{T}}\right)=\Re\left\{\text{A}\cdot\text{a}\cdot\frac{\cos\left(\text{a}\zeta\right)-\sin\left(\text{a}\zeta\right)i}{\exp\left[\text{a}\sigma\right]}\right\}=\text{A}\cdot\text{a}\cdot\frac{\cos\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\tag2$$ 2.$$\Im\left(\text{A}\cdot\text{a}\cdot e^{-\text{a}\cdot\text{T}}\right)=\Im\left\{\text{A}\cdot\text{a}\cdot\frac{\cos\left(\text{a}\zeta\right)-\sin\left(\text{a}\zeta\right)i}{\exp\left[\text{a}\sigma\right]}\right\}=-\text{A}\cdot\text{a}\cdot\frac{\sin\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\tag3$$

Now, for the RHS:

$$\frac{2i}{3\cdot\Re\left(\text{T}\right)}\cdot\left(\omega_0+\text{A}\cdot e^{-\text{a}\cdot\text{T}}\right)=\frac{2i}{3\cdot\sigma}\cdot\left(\omega_0+\text{A}\cdot\frac{\cos\left(\text{a}\zeta\right)-\sin\left(\text{a}\zeta\right)i}{\exp\left[\text{a}\sigma\right]}\right)=$$ $$\frac{2}{3\cdot\sigma}\cdot\left(\omega_0i+\text{A}\cdot\frac{\sin\left(\text{a}\zeta\right)+\cos\left(\text{a}\zeta\right)i}{\exp\left[\text{a}\sigma\right]}\right)\tag4$$

So, we can write:

1.$$\Re\left\{\frac{2i}{3\cdot\Re\left(\text{T}\right)}\cdot\left(\omega_0+\text{A}\cdot e^{-\text{a}\cdot\text{T}}\right)\right\}=\frac{2}{3\cdot\sigma}\cdot\left(\text{A}\cdot\frac{\sin\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\right)\tag5$$ 2.$$\Im\left\{\frac{2i}{3\cdot\Re\left(\text{T}\right)}\cdot\left(\omega_0+\text{A}\cdot e^{-\text{a}\cdot\text{T}}\right)\right\}=\frac{2}{3\cdot\sigma}\cdot\left(\omega_0+\text{A}\cdot\frac{\cos\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\right)\tag6$$

So, we can set up a system of equations:

$$ \begin{cases} \text{A}\cdot\text{a}\cdot\frac{\cos\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}=\frac{2}{3\cdot\sigma}\cdot\left(\text{A}\cdot\frac{\sin\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\right)\\ \\ -\text{A}\cdot\text{a}\cdot\frac{\sin\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}=\frac{2}{3\cdot\sigma}\cdot\left(\omega_0+\text{A}\cdot\frac{\cos\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\right) \end{cases}\tag7 $$

This leads towards, this simplified system of equations:

$$ \begin{cases} \text{a}\cdot\frac{\cos\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}=\frac{2}{3\cdot\sigma}\cdot\frac{\sin\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\\ \\ \frac{\sin\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}=-\frac{1}{\text{A}\cdot\text{a}}\cdot\frac{2}{3\cdot\sigma}\cdot\left(\omega_0+\text{A}\cdot\frac{\cos\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\right) \end{cases}\tag8 $$

So:

$$\text{a}^2\cdot\frac{\cos\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}=-\frac{4}{9\cdot\sigma^2}\cdot\left(\frac{\omega_0}{\text{A}}+\frac{\cos\left(\text{a}\zeta\right)}{\exp\left[\text{a}\sigma\right]}\right)\tag9$$

  • 1
    This is really enlightening, thank you very much! – 2017-01-19
  • 0
    @AlanTuring You're welcome, I'm glad that I could help. But this is not what you're looking forI think, because I used the real and imaginary part of $\text{T}$ apart. – 2017-01-19