My Math Forum Complex numbers question

 Complex Analysis Complex Analysis Math Forum

 November 2nd, 2010, 09:56 AM #1 Member   Joined: Apr 2010 Posts: 91 Thanks: 0 Complex numbers question Which complex numbers satisfy: $z= |z|$? Attempt: $z= |z|$ $x + yi= \sqrt{x^2 + y^2}$ $(x + iy)^2= x^2 + y^2$ I am not sure what else there is to do
 November 2nd, 2010, 12:07 PM #2 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: Complex numbers question Actually, you know that $|z|$ is a real number. Thus, z must be a real number. When does $x= |x|?$ whenever x is a real number?
November 3rd, 2010, 06:15 AM   #3
Member

Joined: Apr 2010

Posts: 91
Thanks: 0

Re: Complex numbers question

Quote:
 Originally Posted by six Actually, you know that $|z|$ is a real number. Thus, z must be a real number. When does $x= |x|?$ whenever x is a real number?
Well x would always equal |x|, since $|x|= \sqrt{x^2} = x$

 November 3rd, 2010, 09:27 AM #4 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: Complex numbers question -1 =\= |-1|
 November 3rd, 2010, 11:11 AM #5 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: Complex numbers question See, the problem is that x^2 doesn't have an inverse! The sqrt is not a function (do you see why?).
November 6th, 2010, 12:09 AM   #6
Member

Joined: Apr 2010

Posts: 91
Thanks: 0

Re: Complex numbers question

Quote:
 Originally Posted by six See, the problem is that x^2 doesn't have an inverse! The sqrt is not a function (do you see why?).
Cause a square root returns 2 values (+ and -)?

 November 6th, 2010, 02:44 AM #7 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: Complex numbers question Precisely. In case you're wondering though, $\sqrt{x^2}$ represents positive values only, since it doesn't have a plus or minus sign in front of it.
 November 16th, 2010, 04:27 PM #8 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: Complex numbers question Six, do you know what sqrt is?
 November 16th, 2010, 05:49 PM #9 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: Complex numbers question Could you please elaborate instead of asking a silly question? Just tell me what's wrong with my reasoning instead, please.
 November 16th, 2010, 06:55 PM #10 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: Complex numbers question Sqrt is certainly a function. It is defined to be the inverse of $x^2$ restricted to $[0,\infty)$. When solving quadratic equations you must use a plus and minus sqrt , but sqrt as a function has nothing to do with +/-

 Tags complex, numbers, question

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post carl17 Algebra 2 October 26th, 2013 10:55 AM Bula Algebra 1 March 27th, 2013 12:48 PM OriaG Algebra 16 August 27th, 2012 02:20 PM dylan182 Complex Analysis 4 December 16th, 2010 06:05 AM OriaG Calculus 16 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top