How $frac{c:e^{jtheta}- a^H x}{|a|_2^2}a rightarrow$ $frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$ for $theta$...












0












$begingroup$


I am reading a paper, there it rewrites this equation



$$frac{c:e^{jtheta}- a^H x}{|a|_2^2}a$$



into



$$frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$$ if $theta$ to be angle of $a^H x$, where



$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$




Sorry, I just don't understand how to derive the latter from former. Please help.




My question in other words, how



$e^{jtheta} = frac{a^H x}{|a^H x|}$ ?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
    $endgroup$
    – Servaes
    Dec 3 '18 at 10:50










  • $begingroup$
    $j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
    $endgroup$
    – learning
    Dec 3 '18 at 10:57


















0












$begingroup$


I am reading a paper, there it rewrites this equation



$$frac{c:e^{jtheta}- a^H x}{|a|_2^2}a$$



into



$$frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$$ if $theta$ to be angle of $a^H x$, where



$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$




Sorry, I just don't understand how to derive the latter from former. Please help.




My question in other words, how



$e^{jtheta} = frac{a^H x}{|a^H x|}$ ?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
    $endgroup$
    – Servaes
    Dec 3 '18 at 10:50










  • $begingroup$
    $j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
    $endgroup$
    – learning
    Dec 3 '18 at 10:57
















0












0








0





$begingroup$


I am reading a paper, there it rewrites this equation



$$frac{c:e^{jtheta}- a^H x}{|a|_2^2}a$$



into



$$frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$$ if $theta$ to be angle of $a^H x$, where



$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$




Sorry, I just don't understand how to derive the latter from former. Please help.




My question in other words, how



$e^{jtheta} = frac{a^H x}{|a^H x|}$ ?






linear-algebra







share|cite|improve this question











$endgroup$




I am reading a paper, there it rewrites this equation



$$frac{c:e^{jtheta}- a^H x}{|a|_2^2}a$$



into



$$frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$$ if $theta$ to be angle of $a^H x$, where



$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$




Sorry, I just don't understand how to derive the latter from former. Please help.




My question in other words, how



$e^{jtheta} = frac{a^H x}{|a^H x|}$ ?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 10:57







learning

















asked Dec 3 '18 at 10:19









learninglearning

566




566












  • $begingroup$
    What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
    $endgroup$
    – Servaes
    Dec 3 '18 at 10:50










  • $begingroup$
    $j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
    $endgroup$
    – learning
    Dec 3 '18 at 10:57




















  • $begingroup$
    What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
    $endgroup$
    – Servaes
    Dec 3 '18 at 10:50










  • $begingroup$
    $j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
    $endgroup$
    – learning
    Dec 3 '18 at 10:57


















$begingroup$
What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
$endgroup$
– Servaes
Dec 3 '18 at 10:50




$begingroup$
What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
$endgroup$
– Servaes
Dec 3 '18 at 10:50












$begingroup$
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
$endgroup$
– learning
Dec 3 '18 at 10:57






$begingroup$
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
$endgroup$
– learning
Dec 3 '18 at 10:57












1 Answer
1






active

oldest

votes


















1












$begingroup$

It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:



A picture is worth a thousand words:



![enter image description here



The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This explains, thank you so much
    $endgroup$
    – learning
    Dec 3 '18 at 10:58











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:



A picture is worth a thousand words:



![enter image description here



The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This explains, thank you so much
    $endgroup$
    – learning
    Dec 3 '18 at 10:58
















1












$begingroup$

It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:



A picture is worth a thousand words:



![enter image description here



The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This explains, thank you so much
    $endgroup$
    – learning
    Dec 3 '18 at 10:58














1












1








1





$begingroup$

It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:



A picture is worth a thousand words:



![enter image description here



The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.






share|cite|improve this answer











$endgroup$



It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:



A picture is worth a thousand words:



![enter image description here



The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 10:54

























answered Dec 3 '18 at 10:43









ServaesServaes

25.7k33996




25.7k33996












  • $begingroup$
    This explains, thank you so much
    $endgroup$
    – learning
    Dec 3 '18 at 10:58


















  • $begingroup$
    This explains, thank you so much
    $endgroup$
    – learning
    Dec 3 '18 at 10:58
















$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58




$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58


















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