Simplifying a product of complex conjugates to a sum












1














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



$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$



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.






algebra-precalculus complex-numbers







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  • 1




    Hint: $2=1+j2pi f + 1-j2pi f$
    – John11
    2 days ago
















1














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



$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$



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.






algebra-precalculus complex-numbers







share|cite|improve this question




















  • 1




    Hint: $2=1+j2pi f + 1-j2pi f$
    – John11
    2 days ago














1












1








1


1





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



$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$



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.






algebra-precalculus complex-numbers







share|cite|improve this question















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



$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$



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.






algebra-precalculus complex-numbers




algebra-precalculus complex-numbers






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited 2 days ago

























asked 2 days ago









Avedis

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396








  • 1




    Hint: $2=1+j2pi f + 1-j2pi f$
    – John11
    2 days ago














  • 1




    Hint: $2=1+j2pi f + 1-j2pi f$
    – John11
    2 days ago








1




1




Hint: $2=1+j2pi f + 1-j2pi f$
– John11
2 days ago




Hint: $2=1+j2pi f + 1-j2pi f$
– John11
2 days ago










1 Answer
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I think you meant:$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$
This is because we can use the fact that $(1+j2pi f)+(1-j2pi f)=2$, to get:
$$frac{1}{1+j2pi f}frac{1}{1-j2pi f}=frac12frac2{(1+j2pi f)(1-j2pi f)}=frac12left(frac{1+j2pi f+1-j2pi f}{(1+j2pi f)(1-j2pi f)}right)\=frac12left(frac1{1-j2pi f}+frac1{1+j2pi f}right)$$
This is generally known as the method of partial fractions.






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  • I saw my error and fixed. thank you for the response!
    – Avedis
    2 days ago











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I think you meant:$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$
This is because we can use the fact that $(1+j2pi f)+(1-j2pi f)=2$, to get:
$$frac{1}{1+j2pi f}frac{1}{1-j2pi f}=frac12frac2{(1+j2pi f)(1-j2pi f)}=frac12left(frac{1+j2pi f+1-j2pi f}{(1+j2pi f)(1-j2pi f)}right)\=frac12left(frac1{1-j2pi f}+frac1{1+j2pi f}right)$$
This is generally known as the method of partial fractions.






share|cite|improve this answer





















  • I saw my error and fixed. thank you for the response!
    – Avedis
    2 days ago
















1














I think you meant:$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$
This is because we can use the fact that $(1+j2pi f)+(1-j2pi f)=2$, to get:
$$frac{1}{1+j2pi f}frac{1}{1-j2pi f}=frac12frac2{(1+j2pi f)(1-j2pi f)}=frac12left(frac{1+j2pi f+1-j2pi f}{(1+j2pi f)(1-j2pi f)}right)\=frac12left(frac1{1-j2pi f}+frac1{1+j2pi f}right)$$
This is generally known as the method of partial fractions.






share|cite|improve this answer





















  • I saw my error and fixed. thank you for the response!
    – Avedis
    2 days ago














1












1








1






I think you meant:$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$
This is because we can use the fact that $(1+j2pi f)+(1-j2pi f)=2$, to get:
$$frac{1}{1+j2pi f}frac{1}{1-j2pi f}=frac12frac2{(1+j2pi f)(1-j2pi f)}=frac12left(frac{1+j2pi f+1-j2pi f}{(1+j2pi f)(1-j2pi f)}right)\=frac12left(frac1{1-j2pi f}+frac1{1+j2pi f}right)$$
This is generally known as the method of partial fractions.






share|cite|improve this answer












I think you meant:$$frac{1}{1+j2pi f}frac{1}{1-j2pi f} = frac{1/2}{1+j2pi f}+frac{1/2}{1-j2pi f}$$
This is because we can use the fact that $(1+j2pi f)+(1-j2pi f)=2$, to get:
$$frac{1}{1+j2pi f}frac{1}{1-j2pi f}=frac12frac2{(1+j2pi f)(1-j2pi f)}=frac12left(frac{1+j2pi f+1-j2pi f}{(1+j2pi f)(1-j2pi f)}right)\=frac12left(frac1{1-j2pi f}+frac1{1+j2pi f}right)$$
This is generally known as the method of partial fractions.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









John Doe

10.4k11135




10.4k11135












  • I saw my error and fixed. thank you for the response!
    – Avedis
    2 days ago


















  • I saw my error and fixed. thank you for the response!
    – Avedis
    2 days ago
















I saw my error and fixed. thank you for the response!
– Avedis
2 days ago




I saw my error and fixed. thank you for the response!
– Avedis
2 days ago


















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