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I got this in classroom

$$Re\left[\frac{1}{j2\pi f_0 RC + 1}Ae^{j2\pi f_0t}\right]$$ $$=Re\left[\frac{1}{\sqrt{4\pi^2 f_0^2 R^2C^2+1}}e^{-j{tan^-1}2\pi f_0 RC}Ae^{j2\pi f_0t}\right]$$

I attend Electrical engineering. $j$ means imaginary number in my major definition instead of $i$. $j=\sqrt{-1}$

I don't know why $\frac{1}{j2\pi f_0 RC + 1}$ can be convert into polar form $\frac{1}{\sqrt{4\pi^2 f_0^2 R^2C^2+1}}e^{-j{tan^-1}2\pi f_0 RC}$. Can someone explain it for me.

If $z$ is a complex number that $z = re^{j\theta}$. So does $\frac1z=\frac1re^{-j\theta}$?

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2 Answers 2

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$$ \frac{1}{j2\pi f_0 RC + 1}= \frac{1}{j2\pi f_0 RC + 1}\frac{1 -j2\pi f_0 RC}{1 - j2\pi f_0 RC}= \frac{1}{1 +4\pi^2 f_0^2 R^2 C^2}(1 -j2\pi f_0 RC)=\cdots $$

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The exponent obeys all the expected rules, such as $1/e^t = e^{-t}$, for any complex number $t$. So indeed, if $z = re^{j\varphi}$, then $1/z = 1/r \cdot 1/e^{j\varphi} = 1/r \cdot e^{-j\varphi}$.

So, if you already understand that $j2\pi f_0 RC + 1 = \sqrt{4\pi^2 f_0^2 R^2 C^2 + 1} \cdot e^{j\tan^{-1}(2\pi f_0 RC)}$, then you apply the above and that is it.

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