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Loser66
 one year ago
Someone explains me, please. Question on attachment.
Loser66
 one year ago
Someone explains me, please. Question on attachment.

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.2I don't get "Equivalently..." why is it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Can the magnitude of a complex number ever be negative ? dw:1440808727196:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2but zw (bar) is not the magnitude.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2well the real part of \(z\overline{w}\) is magnitude of something

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440809022356:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2Question: when a complex number is consider as a negative one?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2complex numbers don't have that sense of binary negative /positive the sense comes from the arguement let me think a bit..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the magnitude of \(z\overline{w}\) equals the real component of \(z\overline{w}\) can we conclude that the imaginary component of \(z\overline{w}\) is 0 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2since the imaginary component is 0, the complex number \(z\overline{w}\) is just as good as a real number and we can compare it with other reals like 0 etc

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Moreover \(z\overline{w}\) identically equals the magnitude of something, so it has to be nonnegative

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2No, not that, like dw:1440809469436:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2By definition, ,magnitude of a complex number is the distance from itself to the origin. Length can never be negative

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440809618870:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2I know, I talk about the vector itself, \(z\overline w\), not \(z\overline w\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\large 5+0i = 5\) is on negative side, so it can never be magnitude of somebody.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2We have \(z\overline w=Re (z\overline w)\) I am ok with that, but Equivalently \(z\overline w\), (this guy is not the magnitude or a Real part of anything, and it is \(\geq 0\) That is what I don't know how to argue.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we have concluded earlier that the imaginary part of \(z\overline{w}\) is 0, yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2therefore \(\Re{(z\overline{w})} = z\overline{w} \)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2ok, got you. hehehe.. yes. Thanks a lot.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the solution in docx kinda sucks... its not detailed and you will need to put together the missing pieces

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440810132148:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2oh, for this part, I know how to fix.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2can you explain me that part is that some identity that you're using or do you know why is it true

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2We have a = b iff a^2 =b^2 Apply to this , we have: \(zw= z w\) iff \( LHS ^2 = RHS^2\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2\(LHS^2 = (zw) \overline{(zw)} =(zw)(\overline z \overline{w})\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2for LHS aren't you simply using the identity \[(ab)^2 = a^22ab+b^2\] ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2\(=z^2z\overline{w}\overline{z}w+w^2\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2The middle terms only \(\overline{z\overline{w}}= \overline{z}w\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2Hence it is \((z\overline{w} +\overline{z\overline{w}})= 2Re(z\overline{w})\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2That is why it becomes as shown.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(=z^2z\overline{w}\overline{z}w+w^2\) can you show me how the middle two terms equal to \(2\Re(z\overline{w})\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i don't get it, let me try reading it again

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Got it! thank you :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2hihihi... I have to say thanks to you :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2looks neat! just writing it again : \[zw^2 = (zw)(\overline{zw}) =(zw)(\overline{z}\overline{w})\\~\\= z^2\color{blue}{w\overline{z} z\overline{w}}+w^2 = z^2\color{blue}{\overline{\overline{w}z}z\overline{w}}+w^2= z^2\color{blue}{2\Re{(\overline{w}z})}+w^2\]
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