Simplifying a product of complex conjugates to a sum












1












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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




    $begingroup$
    Hint: $2=1+j2pi f + 1-j2pi f$
    $endgroup$
    – John11
    Jan 3 at 19:35
















1












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    Hint: $2=1+j2pi f + 1-j2pi f$
    $endgroup$
    – John11
    Jan 3 at 19:35














1












1








1


1



$begingroup$


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











$endgroup$




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













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 19:52







Avedis

















asked Jan 3 at 19:32









AvedisAvedis

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476








  • 1




    $begingroup$
    Hint: $2=1+j2pi f + 1-j2pi f$
    $endgroup$
    – John11
    Jan 3 at 19:35














  • 1




    $begingroup$
    Hint: $2=1+j2pi f + 1-j2pi f$
    $endgroup$
    – John11
    Jan 3 at 19:35








1




1




$begingroup$
Hint: $2=1+j2pi f + 1-j2pi f$
$endgroup$
– John11
Jan 3 at 19:35




$begingroup$
Hint: $2=1+j2pi f + 1-j2pi f$
$endgroup$
– John11
Jan 3 at 19:35










1 Answer
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$begingroup$

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











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$begingroup$

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









$endgroup$













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
















1












$begingroup$

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









$endgroup$













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














1












1








1





$begingroup$

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









$endgroup$



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 Jan 3 at 19:49









John DoeJohn Doe

11.2k11238




11.2k11238












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


















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
















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




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


















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