Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\frac1z=\frac1re^{-j\theta}$?

share|improve this question

2 Answers 2

up vote 0 down vote accepted

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.