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August 30th, 2017, 01:52 PM  #1 
Senior Member Joined: Nov 2015 From: United States of America Posts: 167 Thanks: 21 Math Focus: Calculus and Physics  Modulus and Argument of z
Hello, I am studying complex numbers. I am a little confused since I see if you have $$z=re^{i\theta} $$ I see you can find $r$ by $$ r = z = \sqrt{x^2+y^2}$$ and $$\theta = \arctan(x,y)$$ I have never seen trigonometric functions with more than one variable. Especially in this case where they are separated by a comma as the argument of the inverse trig function. I see the solution for $$\theta =  \frac{\pi}{3}$$ How do you reach this solution? Thanks edit: I get $$ \frac{\pi}{3} $$ if I treat (x,y) like any other coordinate point and then solve it for the angle. I suppose that is the proper method for evaluating these. Last edited by skipjack; August 30th, 2017 at 03:41 PM. 
August 30th, 2017, 02:40 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,942 Thanks: 1009 
the old $\arctan()$ with two arguments trick... This is exactly what you would expect it to be. An $\arctan()$ function that can return the angle in the correct quadrant, given both arguments, instead of just their quotient. 
August 30th, 2017, 03:10 PM  #3 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
$ tan { \theta } = \frac{y}{x} $ $ \theta = tan^{1} ( \frac{y}{x}) $ So writing it as $ \ \ \theta = arctan( x , y) \ \ $ is asking for trouble in my opinion. 
August 30th, 2017, 03:26 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,942 Thanks: 1009  Quote:
With the one argument version the range has a width of $\pi$ With the two argument version the range has a width of $2\pi$ The two argument version is able to distinguish between say $\arctan\left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right) =\dfrac{\pi}{4} \text{ and } \arctan \left( \dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right) =\dfrac{5\pi}{4}$  
August 30th, 2017, 03:38 PM  #5  
Senior Member Joined: Aug 2012 Posts: 1,889 Thanks: 525  Quote:
It's been a while since I've been in school so for all I know this is a math convention now too.  
August 30th, 2017, 03:47 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
$\arctan \left({\large \frac{1}{√2}}, \large\frac{1}{√2}\right) = \large\frac{3\pi}{4}$ The 2argument version used in computer programming would usually be atan2(y, x). 
August 30th, 2017, 04:01 PM  #7  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
However, I can easily see how a beginner may accidentally calculate the wrong quotient because of this ambiguous notation $ (x , y) $ More clearly, they may enter it into the calculator as $ \ \ \frac{x}{y} \ \ $ by mistake Last edited by skipjack; August 30th, 2017 at 04:41 PM.  
August 30th, 2017, 04:42 PM  #8 
Senior Member Joined: Nov 2015 From: United States of America Posts: 167 Thanks: 21 Math Focus: Calculus and Physics 
Thanks for clearing this up folks.

August 30th, 2017, 05:56 PM  #9 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
By definition, tan$\displaystyle \theta$=y/x on the unit circle. arctan (y/x)=$\displaystyle \theta$ What is your definition of arctan x? 

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