Complex Analysis Complex Analysis Math Forum

 August 30th, 2017, 02:52 PM #1 Senior Member   Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 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 04:41 PM. August 30th, 2017, 03:40 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 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. Thanks from SenatorArmstrong August 30th, 2017, 04:10 PM #3 Math Team   Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 $tan { \theta } = \frac{y}{x}$ $\theta = tan^{-1} ( \frac{y}{x})$ So writing it as $\ \ \theta = arctan( x , y) \ \$ is asking for trouble in my opinion. Thanks from SenatorArmstrong August 30th, 2017, 04:26 PM   #4
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 Originally Posted by agentredlum $tan { \theta } = \frac{y}{x}$ $\theta = tan^{-1} ( \frac{y}{x})$ So writing it as $\ \ \theta = arctan( x , y) \ \$ is asking for trouble in my opinion. no. It works perfectly well.

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, 04:38 PM   #5
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 Originally Posted by romsek no. With the two argument version the range has a width of $2\pi$
I think (I could be wrong) that the two-argument arctan is more of a programming convention. I think that in math we'd more likely talk about the principle value of the arctan.

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, 04:47 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,127 Thanks: 2336 $\arctan \left(-{\large \frac{1}{√2}}, -\large\frac{1}{√2}\right) = -\large\frac{3\pi}{4}$ The 2-argument version used in computer programming would usually be atan2(y, x). August 30th, 2017, 05:01 PM   #7
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 Originally Posted by romsek no. It works perfectly well. 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 $\arctan\left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right) =\dfrac{\pi}{4} \text{ and } \arctan \left( \dfrac{-1}{\sqrt{2}}, \dfrac{-1}{\sqrt{2}}\right) =\dfrac{5\pi}{4}$
I don't see a 1 argument version. If you are using a popular scientific calculator, you still need to determine which quadrant you are in.

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 $\ \ \frac{x}{y} \ \$ by mistake Last edited by skipjack; August 30th, 2017 at 05:41 PM. August 30th, 2017, 05:42 PM #8 Senior Member   Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics Thanks for clearing this up folks. August 30th, 2017, 06:56 PM #9 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 By definition, tan$\displaystyle \theta$=y/x on the unit circle. arctan (y/x)=$\displaystyle \theta$ What is your definition of arctan x? Tags argument, modulus Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post asred9 Complex Analysis 7 January 8th, 2016 03:50 PM ach4124 Complex Analysis 1 September 12th, 2015 04:04 PM bilano99 Calculus 2 March 15th, 2013 01:48 PM queenie_n Complex Analysis 2 October 27th, 2012 09:08 AM Hemi08 Abstract Algebra 1 August 20th, 2009 08:54 PM

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