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August 30th, 2017, 01:52 PM   #1
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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?

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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.
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August 30th, 2017, 02:40 PM   #2
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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.
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August 30th, 2017, 03:10 PM   #3
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$ 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.

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August 30th, 2017, 03:26 PM   #4
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Quote:
Originally Posted by agentredlum View Post
$ 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}$
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August 30th, 2017, 03:38 PM   #5
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Quote:
Originally Posted by romsek View Post
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.
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August 30th, 2017, 03:47 PM   #6
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$\arctan \left(-{\large \frac{1}{√2}}, -\large\frac{1}{√2}\right) = -\large\frac{3\pi}{4}$

The 2-argument version used in computer programming would usually be atan2(y, x).
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August 30th, 2017, 04:01 PM   #7
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Quote:
Originally Posted by romsek View Post
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.
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August 30th, 2017, 04:42 PM   #8
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Thanks for clearing this up folks.
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August 30th, 2017, 05:56 PM   #9
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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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