Simplifying a product of complex conjugates to a sum

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.

• Hint: $2=1+j2\pi f + 1-j2\pi f$ – John11 Jan 3 '19 at 19:35

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.