Why multiplying complex current $\hat{I}(\omega)$ with $e^{-i\omega t}$ and taking the real part gives actual current?

Question

In modern electrodynamics by Andrew Zangwill chapter 14, section 14.13.2 an analysis of RLC circuit is shown where Fourier transform of current, EMF, and impedance is used. And equation is $\hat{E}(\omega)=\hat{Z}(\omega)\hat{I}(\omega)$. Then it says in equation 14.148 that the physical current driven by a real electromotive force $\hat{E}(\omega)\cos(\omega t)$ is $$I(t)=\Re[\hat{I}(\omega)\exp(-i\omega t)]=\frac{\hat{E}(\omega)}{\sqrt{R^2+(1/\omega C-\omega L)^2}}\cos(\omega t+\phi)$$ I get that the middle implies the right side but I couldn't figure out why this equal to $I(t)$ which we should get by inverse Fourier transform from $\hat{I}(\omega)$. Any help from anyone is much appreciated.

Answer 1

Go through Phasors.

Phasors are used to represent a time-domain signal $f(t)$ as a frequency domain signal $\hat{f}(\omega)$ so as to solve the integro-differential equations quickly.

We have,

$\exp[\pm i(\omega t+ \phi)] = \cos(\omega t + \phi) \pm i\sin(\omega t + \phi). $

So,

$\cos(\omega t + \phi) = \mathfrak{Re}\{\exp[\pm i(\omega t+ \phi)]\}$
and
$\sin(\omega t + \phi) = \pm\ \mathfrak{Im}\{\exp[\pm i(\omega t+ \phi)]\}$

If we have current, $I(t) = I\cos(\omega t+\phi)$ (where $I =$ amplitude of $I(t) $), we can write

$I(t) = \mathfrak{Re}\{I\exp[\pm i(\omega t+ \phi)]\}$

or, $I(t) = \mathfrak{Re}\{I\exp[\pm i(\omega t+ \phi)]\} = \mathfrak{Re}\{\hat{I}\exp(\pm i\omega t)\}$ where $\hat{I} = I\exp(\pm i\phi)$

Here as cosine function is even, you can consider either the $+$ or $-$ sign.

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