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I found the following simplification but unsure how it to derive it.

$$\frac{1}{1+j2\pi f}\frac{1}{1-j2\pi f} = \frac{1/2}{1+j2\pi f}+\frac{1/2}{1-j2\pi f}$$

The latter form makes taking the Fourier transform much easier than the former. My initial thought was to condence the demonominator into to the form of $(a+b)^2$ but that form doesn't help me.

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    $\begingroup$ Hint: $2=1+j2\pi f + 1-j2\pi f$ $\endgroup$ – John11 Jan 3 '19 at 19:35
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I think you meant:$$\frac{1}{1+j2\pi f}\frac{1}{1-j2\pi f} = \frac{1/2}{1+j2\pi f}+\frac{1/2}{1-j2\pi f}$$ This is because we can use the fact that $(1+j2\pi f)+(1-j2\pi f)=2$, to get: $$\frac{1}{1+j2\pi f}\frac{1}{1-j2\pi f}=\frac12\frac2{(1+j2\pi f)(1-j2\pi f)}=\frac12\left(\frac{1+j2\pi f+1-j2\pi f}{(1+j2\pi f)(1-j2\pi f)}\right)\\=\frac12\left(\frac1{1-j2\pi f}+\frac1{1+j2\pi f}\right)$$ This is generally known as the method of partial fractions.

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  • $\begingroup$ I saw my error and fixed. thank you for the response! $\endgroup$ – Avedis Jan 3 '19 at 19:52

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