Why Does A Equal the Reciprocal of the Modulus in This Complex ODE Solution?

[gdbb]

I'm trying to solve the real part of the ODE

$\tilde{y}' + k\tilde{y} = ke^{i \omega t}$

in polar form, and I'm running into a problem. I understand up until here:

$\frac {1} {1 + i(\frac {\omega} {k})} = Ae^{-i \phi}$

which I can see is equal to

$A(cos(- \phi) + isin(- \phi))$

and that is where I get stuck. Apparently, with the use of right triangle trig in the complex domain, this is true:

$A = \frac {1} {\sqrt{1 + (\frac {\omega} {k})^2}}$

and I cannot for the life of me understand why A is equal to the reciprocal of the absolute value of the modulus, rather than just the absolute value of the modulus itself.

Please help!

 

[Redbelly98]

As you said, A is equal to the absolute value of the modulus, which is $\frac{1}{\sqrt{1+ ( \frac{ \omega }{h})^2}}$ as written.

 

[gdbb]

Hmm, I guess I just don't understand the complex domain, then. Shouldn't the modulus be equal to

$\sqrt{Im^2 + Re^2}$

for which $$Im$$ is the imaginary component and $$Re$$ is the real component of the complex number? I know that

$$\arg{\alpha} = -\arg{\frac {1} {|\alpha|}}$$

but, if that leads to the answer here, I don't see the connection, as the argument function deals with angles, and the modulus is the magnitude, not the angle.

 

[Redbelly98]

Hmm, I guess I just don't understand the complex domain, then. Shouldn't the modulus be equal to

$\sqrt{Im^2 + Re^2}$

for which $$Im$$ is the imaginary component and $$Re$$ is the real component of the complex number?

Yes, that's right.

So, what are the real and imaginary parts of $\frac {1} {1 + i(\frac {\omega} {k})}$ ?

 

[gdbb]

Oh! We multiply by the complex conjugate. Let me see if I can work this out now...

$\begin{align*} <br /> z = \frac {1} {1 + i( \frac {\omega} {k})} &= \frac {1 - i(\frac {\omega} {k})} {1 + (\frac {\omega} {k})^2}\\<br /> <br /> &= \frac {1} {1+(\frac {\omega} {k})^2} - \frac {i(\frac {\omega} {k})} {1 + (\frac {\omega} {k})^2} <br /> <br /> \end{align*}$

Rewriting and taking the real part of that...

$\begin{align*}<br /> <br /> z \bar{z} = |z|^2 &= \frac {1} {1 + (\frac {\omega} {k})^2} \\<br /> <br /> \Rightarrow |z| &= \frac {1} {\sqrt{1 + (\frac {\omega} {k})^2}}<br /> <br /> \end{align*}$

Hmm. I feel like I just went around in a circle. I mean, this work makes sense, but I still don't understand why the amplitude A is equal to the reciprocal of the modulus, rather than just the modulus itself.

 

[Redbelly98]

It is equal to the modulus itself, which as I said before is $\frac{1}{\sqrt{1+ ( \frac{ \omega }{h})^2}}$. And you have just proved it.

 

[gdbb]

Alright, I think I understand now. Thanks!