Can this complex equation be solved for a real number result?

Question

I have the following equation:

$v(t) =\frac{-I_0}{C}\frac{e^{-at}-e^{-bt}}{b-a}$

$a=\frac{L+\sqrt{L^{2}-16CLR^{2}}}{4CLR}$

$b=\frac{L-\sqrt{L^{2}-16CLR^{2}}}{4CLR}$

$C$, $L$, $R$, $I_o$ and $t$ are all real positive numbers.

In the condition where $C > \frac{m}{16R^2}$ this results in a square root of a negative in $a$ and $b$. I am trying to figure out how to deal with this. This is solving for voltage, so I am told if I have done everything right, the imaginary component should cancel out and I should get a real result, but I can't seem to do that.

In the condition where the square root is negative we get:

$a = \frac{L}{4CLR}+\frac{i\sqrt{|L^2-16CLR^2|}}{4CLR}$

$b = \frac{L}{4CLR}-\frac{i\sqrt{|L^2-16CLR^2|}}{4CLR}$

If we say:

$c = \frac{1}{4CR}$

$d = \frac{\sqrt{|L^2-16CLR^2|}}{4CLR}$

Then:

$a = c + id$

$b = c-id$

Then our equation becomes:

$v(t) =\frac{-I_0}{C}\frac{e^{-at}-e^{-bt}}{b-a}$

$v(t) =\frac{-I_0}{C}\frac{e^{-(c+id)t}-e^{-(c-id)t}}{(c-id)-(c+id)}$

$v(t) =\frac{-I_0}{C}\frac{e^{-ct-idt}-e^{-ct+idt}}{-2id}$

$v(t) =\frac{-I_0}{C}\frac{e^{-ct}/e^{idt}-e^{-ct}*e^{idt}}{-2id}$

$v(t) =\frac{-I_0 e^{-ct}}{Cd}\frac{e^{-idt}-e^{idt}}{-2i}$

$v(t) =\frac{-I_0 e^{-ct}}{Cd}\frac{cos(-dt) + isin(-dt) - cos(dt) - isin(dt)}{-2i}$

$v(t) =\frac{-I_0 e^{-ct}}{Cd}\frac{sin(-dt) - sin(dt)}{-2}$

$v(t) =\frac{-I_0}{Cde^{ct}}sin(dt)$

Edit: This appears to be the correct solution as per answer already provided and accepted. Thanks.

Answer 1

No, you don't have to go anywhere. If you have continued simplifying the expression you got for $v\left(t\right)$ as shown below, you would have arrived at your goal without any difficulty, i.e. $$v\left(t\right)=-\frac{I_0}{C}\frac{1}{de^{ct}}\frac{e^{idt}-e^{-idt}}{2i}=-\frac{I_0}{C}\frac{1}{de^{ct}}\sin\left(dt\right).$$

Answer 2

$$\frac{e^{(-1/\tau+i\omega)t}-e^{(-1/\tau-i\omega)t}}{(-1/\tau+i\omega)-(-1/\tau-i\omega)}=e^{-t/\tau}\frac{e^{i\omega t}-e^{-i\omega t}}{2i\omega}=e^{-t/\tau}\sin\omega t.$$

This is a damped sinusoid.