Transfer function to phase transfere

[Rectifier]

Hello!
This is problem is a part of a bigger problem which I solved and came to a formula which is correct.

This is the equation for one transfer function. The next thing I would like to find out here is when arg(H(jw)) is -90 or 90 degrees but I get stuck.

Transfer function:
$H(jw) = \frac{R}{R(1-w^2LC)+jwL}$This is how I proceeded till I got stuck.

$H(jw) = \frac{R}{R(1-w^2LC)+jwL} \\ H(jw) = \frac{R}{\sqrt{(R(1-w^2LC))^2+(wL)^2}e^{jarctan( \frac{wL}{R(1-w^2LC)})}} \\ H(jw) = \frac{ R }{ \sqrt{ (R(1-w^2LC))^2+(wL)^2} } e^{-jarctan( \frac{wL}{R(1-w^2LC)})} \\$

Then we want to know where the argument is -90 or 90 degrees.

$90=-jarctan( \frac{wL}{R(1-w^2LC)})$

Here is the step where I get stuck. Could you please help me out?
Thanks in advance!

 

[M Quack]

Where does the "j" on the right hand side of the last line come from?

 

[Rectifier]

Where does the "j" on the right hand side of the last line come from?

Thank you for your reply!

From j here:
$e^{ -jarctan( \frac{wL}{R(1-w^2LC)})}$

But I guess it shouldn't be there. Since I an looking for the phase.

What about $90=-arctan( \frac{wL}{R(1-w^2LC)})$ on the last line then :) ?

EDIT: the problem I have is that 90=-arctan(x) has no solutions :,(. But the solution in my book is $w= \frac{1}{\sqrt{LC}}$. When I try to insert the solution from the book inside the equation I have gives something that is not defined:
$90=-arctan( \frac{\frac{1}{\sqrt{LC}}L}{R(1-(\frac{1}{\sqrt{LC}})^2LC)}) \\ 90= -arctan( \frac{\frac{1}{\sqrt{LC}}L}{R(1-(\frac{1}{1}))}) \\ 90= -arctan( \frac{\frac{1}{\sqrt{LC}}L}{0})$

 

[M Quack]

What does the tangent function look like, in particular, what is the value of tan(90 deg) and tan(-90 deg)?

The mathematically correct way would be to calculate the solution for +/-(90 + epsilon), and then take the limit epsilon-> 0.