Talk:Complex conjugate

message: The article should probably be converted from asterisk notation (a*) to overline notation (\overline{a}) to avoid confusion with the conjugate transpose.  I haven't seen anything but conjugation use overline notation, has anyone else? -FunnyMan 18:02, Oct 23, 2004 (UTC)
:I strongly second this proposal: overline is the standard notation in complex analysis and geometry.--FpWl 18:18, 29 Apr 2005 (UTC)
replies: text: I also endorse this proposal, if nothing else it helps the article agree with the complex numbers entry.  I went ahead and made the change. Thenub314 21:07, 19 July 2006 (UTC)
thread_title:  Asterisks versus overlines 

message: I removed "if w is non-zero" from the following: 
: \left({\frac{z}{w}}\right)^* = \frac{z^*}{w^*} if w is non-zero
because I am almost certain that the equality still holds when w is zero. Even though dividing by zero is not desirable in many cases, it is still mathematically *defined* as "undefined" (in most number systems). Please correct me if i'm wrong. Fresheneesz 08:40, 8 March 2006 (UTC)
thread_title: "if w is non-zero" removed (small item)

message: Hey -- I removed the template, since I did the job. If I missed anything, just let me know! --NicApicella 16:06, 13 March 2006 (UTC)
thread_title:  LaTeX template removed 

message: I think that the initial definition of z ought to be in a bigger font, but short of doing something hacky, I don't know how to make it so. --anon
: Done. Oleg Alexandrov (talk) 18:55, 22 August 2006 (UTC)
thread_title: typography

message: Its obviously NOT idempotent because in particular '\overline{\overline{\overline{z}}} \ne z \!\ ''
I've never heard this usage of the term idempotence. Could someone please point me to some references? (Because I haven't found any.) I think this is confusing (if not bogus), and should be removed.
(I know this is stated in the idempotence article, too, and I think it should be removed from there as well.) --Matt Kovacs (talk) 17:08, 24 January 2009 (UTC)
thread_title:  "Idempotence", huh? 

message: I'm sorry, I still don't get it.  http://xkcd.com/849/  --204.246.229.130 (talk) 15:59, 20 January 2011 (UTC)
replies: text: Explaining a joke usually means to destroy it, but … In theoretical physics, the wavefunction is a complex function. If you multiply any complex number by its complex conjugate, the result is a real number (namely the magnitude squared). The same is true for the wavefunction: multiplying it with its complex conjugate turns it into a real-valued probability density, which actually has physical meaning. (The complex wavefunction itself arguably doesn’t – it’s merely a theoretical construct and cannot be measured in a physical experiment.) I guess the funny part comes from the play with words: on the one hand “real” means it’s a real (and not a complex) number. On the other hand, “real” means we’re now talking about hands-on physics instead of elusive theoretical calculations. Physical observables are always real values, and they can – in principle – be measured. Complex values cannot. Experimental physicists might say that complex numbers only make sense in mathematics, which they’d consider “merely” an auxiliary discipline. Compare http://xkcd.com/435/ including its hover-tooltip comment. --78.48.0.203 (talk) 21:49, 14 September 2011 (UTC)
thread_title:  xkcd 

message: Where did the concept of complex conjugates came from?  It would be great if someone  with that knowledge would make an addition to the article.  — Preceding unsigned comment added by 101.208.119.211 (talk) 08:39, 3 March 2012 (UTC)
thread_title:  History 

message: I don't get the follwing statement from the article:
:: "\overline{(zw)} = \overline{w}\; \overline{z} \!\  (note the reversed arguments if z and w don't commute)"
For complex numbers, multiplication is always commutative so in no way z and w will not commute. Maybe this may be true for non-scalar or matrix algerba, but for complex numbers, the order of operands is unimportant.89.137.186.101 (talk) 20:05, 6 September 2012 (UTC)
thread_title:  (Complex) multiplication is always commutative 

message: No it is not i.e. nonsense; introduced by this edit 14-Apr-2015 Nxavar.
Fixing it means resplicing/reverting the relevant rearranged material - see the linked edit for the original paragraph.  — Preceding unsigned comment added by 2.99.166.53 (talk) 15:23, 7 June 2015 (UTC)
replies: text: You are right, it is total gibberish.  (The author of that sentence seemed to confuse equality with "can be interchanged by a field isomorphism," or something.) I have removed the offending sentence. --JBL (talk) 15:33, 7 June 2015 (UTC)
thread_title:  "... a complex number is equal to its complex conjugate if its imaginary part is [...] -i." 

message: The article states that
\begin{align}
               \overline{z^n} &= \left(\overline{z}\right)^n,\quad \forall n \in \mathbb{Z} \\
\end{align}
, meaning it's only valid for integer n.
However, empirically testing with Wolfram Mathematica shows all real n satisfy the equation. Has it just never been proven so, or is it a mistake?
31.217.16.25 (talk) 18:15, 9 September 2015 (UTC)
replies: text: The issue is complications from multiple values: when you raise complex numbers to non-integer powers, you get many values back, and so it's not trivial to explain what you mean by an equality of the sort in question if n isn't an integer -- choosing a branch of the logarithm, etc. etc.  See Exponential_function#Complex_plane for more about this.  Probably there's no consistent way to choose a branch of the logarithm to make it true once and for all for all z and (complex) n, though maybe with natural choices of branch cuts it is true for all real n.  --JBL (talk) 18:33, 9 September 2015 (UTC)
thread_title:  Power of conjugate and conjugate of power