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March 28, 2024, 08:23:12 pm

Author Topic: Converting Cartesian Form to Modulas-Argument Form  (Read 884 times)  Share 

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dream chaser

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Converting Cartesian Form to Modulas-Argument Form
« on: February 22, 2020, 07:58:59 pm »
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Hi Guys,

Could someone please show me how to convert 1 + tanθ i into modulas-argument form, 0 < θ < π/2

The answer is secθ cisθ. I know how to get the modulas part. Only the argument part I am stuck with.

All help will be much appreciated. Thanks

fun_jirachi

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Re: Converting Cartesian Form to Modulas-Argument Form
« Reply #1 on: February 22, 2020, 08:47:45 pm »
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There are probably two big ways of doing this! :)

Method 1:


Method 2:
Construct a triangle in the first quadrant with base 1, height tan θ, (and therefore hypotenuse sec θ!). The enclosed angle will also be θ, since the tan ratio is literally tan θ/1 ! This triangle thus gives you the modulus and argument almost straight away, without any algebra.

Hope this helps :)
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dream chaser

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Re: Converting Cartesian Form to Modulas-Argument Form
« Reply #2 on: February 22, 2020, 08:52:35 pm »
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There are probably two big ways of doing this! :)

Method 1:


Method 2:
Construct a triangle in the first quadrant with base 1, height tan θ, (and therefore hypotenuse sec θ!). The enclosed angle will also be θ, since the tan ratio is literally tan θ/1 ! This triangle thus gives you the modulus and argument almost straight away, without any algebra.

Hope this helps :)

Thanks fun_jirachi for the help. Much appreciated.

Doing the second method you said, θ=tan^(-1)(tanθ/1) would equal θ? Is it because tan and tan^(-1) cancel each other out or something?

fun_jirachi

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Re: Converting Cartesian Form to Modulas-Argument Form
« Reply #3 on: February 22, 2020, 10:13:47 pm »
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Just to answer your follow-up:

To clarify what I said, here's a diagram:
Spoiler
Basically, we are given the side lengths parallel to the coordinate axes, 1 and tan θ. We can fill in the length of the hypotenuse using Pythagoras' theorem/trigonometric identity. Recall that for any number x+iy on the complex plane that the tangent of its argument will just be y/x. Hence, the tangent of the angle the hypotenuse makes with the real axis (opposite/adjacent !!) will be tan θ/1 as given. This way of thinking eliminates the requirement of using arctan - which really just simplifies things.

To answer the additional question:
You can't always assume that arctan and tan will cancel each other out - recall that arctan(tan x) = x if and only if \(x \in \left(-\frac{\pi}{2}, \ \frac{\pi}{2}\right)\). Here, this assumption is okay because the domain we're given is a subset of the domain for which this identity holds.

Something to also note is that this problem is simplified a lot because of the assumption that θ lies in the first quadrant. This means that every trigonometric ratio is positive and we don't have to worry about minus signs or anything. Basically, it means we don't really have to consider the domain of any of the functions to a rigorous degree.
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dream chaser

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Re: Converting Cartesian Form to Modulas-Argument Form
« Reply #4 on: February 22, 2020, 10:29:27 pm »
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Just to answer your follow-up:

To clarify what I said, here's a diagram:
Spoiler
Basically, we are given the side lengths parallel to the coordinate axes, 1 and tan θ. We can fill in the length of the hypotenuse using Pythagoras' theorem/trigonometric identity. Recall that for any number x+iy on the complex plane that the tangent of its argument will just be y/x. Hence, the tangent of the angle the hypotenuse makes with the real axis (opposite/adjacent !!) will be tan θ/1 as given. This way of thinking eliminates the requirement of using arctan - which really just simplifies things.

To answer the additional question:
You can't always assume that arctan and tan will cancel each other out - recall that arctan(tan x) = x if and only if \(x \in \left(-\frac{\pi}{2}, \ \frac{\pi}{2}\right)\). Here, this assumption is okay because the domain we're given is a subset of the domain for which this identity holds.

Something to also note is that this problem is simplified a lot because of the assumption that θ lies in the first quadrant. This means that every trigonometric ratio is positive and we don't have to worry about minus signs or anything. Basically, it means we don't really have to consider the domain of any of the functions to a rigorous degree.

Thank you so much. Really appreciate the detailed explanation.  :)