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

  • Thread starter gdbb
  • Start date
  • Tags
    Complex
In summary, the real part of the ODE equation is equal to the modulus, and the modulus is equal to the reciprocal of the absolute value of the modulus.
  • #1
gdbb
51
0
I'm trying to solve the real part of the ODE

[itex]\tilde{y}' + k\tilde{y} = ke^{i \omega t}[/itex]

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

[itex]\frac {1} {1 + i(\frac {\omega} {k})} = Ae^{-i \phi}[/itex]

which I can see is equal to

[itex]A(cos(- \phi) + isin(- \phi)) [/itex]

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

[itex]A = \frac {1} {\sqrt{1 + (\frac {\omega} {k})^2}}[/itex]

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!
 
Physics news on Phys.org
  • #2
As you said, A is equal to the absolute value of the modulus, which is [itex]\frac{1}{\sqrt{1+ ( \frac{ \omega }{h})^2}}[/itex] as written.
 
  • #3
Hmm, I guess I just don't understand the complex domain, then. Shouldn't the modulus be equal to

[itex] \sqrt{Im^2 + Re^2}[/itex]

for which [tex]Im[/tex] is the imaginary component and [tex]Re[/tex] is the real component of the complex number? I know that

[tex] \arg{\alpha} = -\arg{\frac {1} {|\alpha|}}[/tex]

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.
 
  • #4
ebob said:
Hmm, I guess I just don't understand the complex domain, then. Shouldn't the modulus be equal to

[itex] \sqrt{Im^2 + Re^2}[/itex]

for which [tex]Im[/tex] is the imaginary component and [tex]Re[/tex] is the real component of the complex number?
Yes, that's right.

So, what are the real and imaginary parts of [itex] \frac {1} {1 + i(\frac {\omega} {k})} [/itex] ?
 
  • #5
Oh! We multiply by the complex conjugate. Let me see if I can work this out now...

[itex] \begin{align*}
z = \frac {1} {1 + i( \frac {\omega} {k})} &= \frac {1 - i(\frac {\omega} {k})} {1 + (\frac {\omega} {k})^2}\\

&= \frac {1} {1+(\frac {\omega} {k})^2} - \frac {i(\frac {\omega} {k})} {1 + (\frac {\omega} {k})^2}

\end{align*}[/itex]

Rewriting and taking the real part of that...

[itex] \begin{align*}

z \bar{z} = |z|^2 &= \frac {1} {1 + (\frac {\omega} {k})^2} \\

\Rightarrow |z| &= \frac {1} {\sqrt{1 + (\frac {\omega} {k})^2}}

\end{align*}[/itex]

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.
 
  • #6
It is equal to the modulus itself, which as I said before is [itex]\frac{1}{\sqrt{1+ ( \frac{ \omega }{h})^2}}[/itex]. And you have just proved it.
 
  • #7
Alright, I think I understand now. Thanks!
 

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

What is the difference between simple and complex trigonometry?

Simple trigonometry involves the study of basic trigonometric functions, such as sine, cosine, and tangent, and their properties. Complex trigonometry, also known as advanced trigonometry, extends these concepts to include more complex functions, such as inverse trigonometric functions, hyperbolic functions, and complex numbers.

How is trigonometry used in solving ordinary differential equations (ODEs)?

Trigonometric functions can be used to represent and solve differential equations involving periodic phenomena, such as oscillations and vibrations. They can also be used to transform non-constant coefficients in a differential equation into constant coefficients, making it easier to solve.

What is the relationship between trigonometry and calculus?

Calculus, specifically integral and differential calculus, is closely related to trigonometry. Trigonometric functions are used to solve many calculus problems, and knowledge of calculus is essential for understanding more advanced trigonometric concepts.

What is the unit circle and how is it used in complex trigonometry?

The unit circle is a circle with a radius of 1 unit, centered at the origin of a Cartesian coordinate system. It is used in complex trigonometry as a visual representation of the relationships between trigonometric functions and angles. The coordinates of points on the unit circle can also be used to evaluate trigonometric functions.

What are some real-life applications of complex trigonometry and ODEs?

Complex trigonometry and ODEs have many practical applications in fields such as physics, engineering, and astronomy. They can be used to model and analyze various phenomena, such as the motion of objects, electrical circuits, and fluid flow. They are also used in the design and optimization of structures, machines, and systems.

Similar threads

Back
Top