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

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.

• You want to solve for $T$? – Simply Beautiful Art Jan 18 '17 at 18:27
• @SimpleArt Yesss! – Von Neumann Jan 18 '17 at 18:28
• :-( Tried Lambert W function, but no matter what, the $\Re(T)$ is being a pain. – Simply Beautiful Art Jan 18 '17 at 18:34
• @SimpleArt Indeed, is that term that makes everything strange! – Von Neumann Jan 18 '17 at 18:34
• have you tried to seperate real and imaginary part? – tired Jan 18 '17 at 20:00

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:

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

• This is really enlightening, thank you very much! – Von Neumann Jan 19 '17 at 10:11
• @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. – Jan Jan 19 '17 at 10:15