My Math Forum Modulus and Argument of z

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 August 30th, 2017, 01:52 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 03:41 PM.
 August 30th, 2017, 02:40 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 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, 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. Thanks from SenatorArmstrong
August 30th, 2017, 03:26 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, 03:38 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, 03:47 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2222 $\arctan \left(-{\large \frac{1}{√2}}, -\large\frac{1}{√2}\right) = -\large\frac{3\pi}{4}$ The 2-argument version used in computer programming would usually be atan2(y, x).
August 30th, 2017, 04:01 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 04:41 PM.

 August 30th, 2017, 04:42 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, 05:56 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?

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