- #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!
[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!