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







share|cite|improve this question




















  • 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













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked 2 days ago









Avedis

396




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
1






active

oldest

votes


















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060912%2fsimplifying-a-product-of-complex-conjugates-to-a-sum%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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














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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060912%2fsimplifying-a-product-of-complex-conjugates-to-a-sum%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

An IMO inspired problem

Image
14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
Read more

Management

Image
Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
Read more

Investment

Image
Investment From Wikipedia, the free encyclopedia Jump to navigation Jump to search This article is about investment in finance. For investment in macroeconomics, see Investment (macroeconomics). For other uses, see Investment (disambiguation). "Invest" redirects here. For the term in meteorology, see Invest (meteorology). This article needs additional citations for verification . Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. ( February 2017 ) (Learn how and when to remove this template message) In general, to invest is to distribute money in the expectation of some benefit in the future – for example, investment in durable goods, in real estate by the service industry, in factories for manufacturing, in product development, and in research and development. However, this article focuses specifically on investment in financial
Read more