Someone explains me, please. Question on attachment.

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Someone explains me, please. Question on attachment.

Mathematics
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I don't get "Equivalently..." why is it?

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Can the magnitude of a complex number ever be negative ? |dw:1440808727196:dw|
No
but zw (bar) is not the magnitude.
well the real part of \(z\overline{w}\) is magnitude of something
|dw:1440809022356:dw|
Question: when a complex number is consider as a negative one?
complex numbers don't have that sense of binary negative /positive the sense comes from the arguement let me think a bit..
the 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 ?
Yes, I think so
since 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
Moreover \(z\overline{w}\) identically equals the magnitude of something, so it has to be nonnegative
No, not that, like |dw:1440809469436:dw|
By definition, ,magnitude of a complex number is the distance from itself to the origin. Length can never be negative
|dw:1440809618870:dw|
I know, I talk about the vector itself, \(z\overline w\), not \(|z\overline w|\)
\(\large -5+0i = -5\) is on negative side, so it can never be magnitude of somebody.
We 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.
we have concluded earlier that the imaginary part of \(z\overline{w}\) is 0, yes ?
Yes
therefore \(\Re{(z\overline{w})} = z\overline{w} \)
ok, got you. hehehe.. yes. Thanks a lot.
np
the solution in docx kinda sucks... its not detailed and you will need to put together the missing pieces
|dw:1440810132148:dw|
oh, for this part, I know how to fix.
can you explain me that part is that some identity that you're using or do you know why is it true
We have |a| = |b| iff |a|^2 =|b|^2 Apply to this , we have: \(||z|-|w||= |z -w|\) iff \( LHS ^2 = RHS^2\)
got that part
\(LHS^2 = (|z|-|w|) \overline{(|z|-|w|)} =(|z|-|w|)(|\overline z| -\overline{|w|})\)
for LHS aren't you simply using the identity \[(a-b)^2 = a^2-2ab+b^2\] ?
\(=|z|^2-|z|\overline{|w|}-\overline{|z|}|w|+|w|^2\)
The middle terms only \(\overline{z\overline{w}}= \overline{z}w\)
Hence it is \(-(z\overline{w} +\overline{z\overline{w}})= -2Re(z\overline{w})\)
That is why it becomes as shown.
|dw:1440810939082:dw|
I did
\(=|z|^2-|z|\overline{|w|}-\overline{|z|}|w|+|w|^2\) can you show me how the middle two terms equal to \(2\Re(z\overline{w})\)
i don't get it, let me try reading it again
|dw:1440811335931:dw|
Got it! thank you :)
|dw:1440811409420:dw|
hihihi... I have to say thanks to you :)
looks neat! just writing it again : \[|z-w|^2 = (z-w)(\overline{z-w}) =(z-w)(\overline{z}-\overline{w})\\~\\= |z|^2\color{blue}{-w\overline{z} -z\overline{w}}+|w|^2 = |z|^2\color{blue}{-\overline{\overline{w}z}-z\overline{w}}+|w|^2= |z|^2\color{blue}{-2\Re{(\overline{w}z})}+|w|^2\]
Thank you. :)

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