Why Is My Argument Calculation for Complex Number Incorrect?

  • #1
MatinSAR
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Homework Statement
Stated below.
Relevant Equations
Complex algebra.
Question 1: Find the modulus and argument of ##z=-\sin \frac {\pi}{8}-i\cos \frac {\pi}{8}##.
The modulus is obviously 1. I can't prove that the argument is ##\frac {-5\pi} {8}##. I think ##\frac {-5\pi} {8}## is not correct ...

What I've done:
$$\tan \theta=\cot \frac {\pi}{8}$$$$\tan \theta=\dfrac {1+\cos \frac {\pi}{4}}{\sin \frac {\pi}{4}}$$$$\tan \theta=1+ \sqrt 2$$$$\theta=\arctan (1+ \sqrt 2) +2k\pi$$
I can't find out why it should be equal to ##\frac {-5\pi} {8}##.

Question 2: Calculate ##(-1+i\sqrt 3)^{60} ##.
We first write it in polar form ##re^{i\theta}##.
$$r=2$$$$\tan \theta = -\sqrt 3 $$$$ \theta = \dfrac {2\pi}{3}$$
We have:
$$ (2e^{i(\dfrac {2\pi}{3})})^{60} =2^{60}e^{i(40\pi)}=2^{60}(\cos 40\pi + i\sin 40\pi)=2^{60}$$
Am I right?

Many thanks.
 
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  • #2
I've checked 2nd question using Microsoft Copilot. It was correct. I did not expect that it can calculate things like this ...
 
  • #3
MatinSAR said:
it can calculate things like this

And it can also calculate ##\arctan\bigl (-\cos({\pi\over 8})/-\sin({\pi\over 8})\bigr ) ## I suppose

But a simple sketch of ##z## on the unit circle will help you better :smile:

##\ ##
 
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  • #4
For another approach, which I think is simpler, try to relate ##z=-\sin(\frac {\pi}{8}) -\cos(\frac {\pi}{8})## to ##e^{i\frac{\pi}{8}} = \cos(\frac {\pi}{8}) +i \sin(\frac {\pi}{8})##.
 
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  • #5
For (1), note that [itex]-z = \sin \frac\pi 8 + i\cos \frac \pi 8[/itex].
Note also that [tex]\begin{split} \cos \theta &= \sin (\frac12\pi - \theta) \\
\sin \theta &= \cos(\frac12\pi - \theta).\end{split}[/tex] Hence [itex]\sin \frac\pi 8 + i \cos \frac \pi 8 = \cos \frac {3 \pi} 8 + i \sin \frac {3\pi}8[/itex].
 
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  • #6
BvU said:
And it can also calculate ##\arctan\bigl (-\cos({\pi\over 8})/-\sin({\pi\over 8})\bigr ) ## I suppose

But a simple sketch of ##z## on the unit circle will help you better :smile:

##\ ##
I've asked first question but the answer was unreadable since my android device doesn't support latex.
FactChecker said:
For another approach, which I think is simpler, try to relate ##z=-\sin(\frac {\pi}{8}) -\cos(\frac {\pi}{8})## to ##e^{i\frac{\pi}{8}} = \cos(\frac {\pi}{8}) +i \sin(\frac {\pi}{8})##.
Thanks for your idea. I'll try.
pasmith said:
For (1), note that [itex]-z = \sin \frac\pi 8 + i\cos \frac \pi 8[/itex].
Note also that [tex]\begin{split} \cos \theta &= \sin (\frac12\pi - \theta) \\
\sin \theta &= \cos(\frac12\pi - \theta).\end{split}[/tex] Hence [itex]\sin \frac\pi 8 + i \cos \frac \pi 8 = \cos \frac {3 \pi} 8 + i \sin \frac {3\pi}8[/itex].
Great! I've understand. Many thanks.
 
  • #7
MatinSAR said:
I can't find out why it should be equal to ##\frac {-5\pi} {8}##.
The trig functions are not 1:1 on the range ##[0, 2\pi]##, so you must take note of what quadrant the number is in. With this step you have already lost that information:
MatinSAR said:
What I've done:
$$\tan \theta=\cot \frac {\pi}{8}$$
 
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  • #8
MatinSAR said:
I've asked first question but the answer was unreadable since my android device doesn't support latex.

Thanks for your idea. I'll try.

Great! I've understand. Many thanks.
I have an Android too, and no problem with Latex. As I understand, the rendering is done in PF itself, not your browser.
 
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  • #9
PeroK said:
The trig functions are not 1:1 on the range ##[0, 2\pi]##, so you must take note of what quadrant the number is in. With this step you have already lost that information:
I'm not sure If I understand your point well. I know from it's cartesian form that it is in 3rd quadrant.
Can't I change that equation to ##\tan \theta = \cot (\pi + \dfrac {\pi} {8})##?
WWGD said:
I have an Android too, and no problem with Latex. As I understand, the rendering is done in PF itself, not your browser.
I don't have problem in PF. In Bing app it doesn't render correctly.
 
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  • #10
MatinSAR said:
I've asked first question but the answer was unreadable since my android device doesn't support latex.
here's another crutch :rolleyes:

##\ ##
 
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  • #11
MatinSAR said:
I'm not sure If I understand your point well.
I know from it's cartesian form that it is in 3rd quadrant. Can't I change that equation to :
$$\tan \theta = \cot (\pi + \dfrac {\pi} {8}) $$

I don't have problem in PF. In Bing app it doesn't render correctly.
It doesn't matter. You start with a specific complex number, but when you calculate ##\tan \theta## you have two possible complex numbers that satisfy that equation. You can never recover ##z## from that equation without additional information.

The complex plane/argand diagram is invaluable. And you should use it as much as possible.
 
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  • #12
BvU said:
Now it is clear. Thanks for your help @BvU ...
I was using google calculator and it didn't gave exact value ... I will use wolframalpha.com from now.
PeroK said:
It doesn't matter. You start with a specific complex number, but when you calculate ##\tan \theta## you have two possible complex numbers that satisfy that equation. You can never recover ##z## from that equation without additional information.

The complex plane/argand diagram is invaluable. And you should use it as much as possible.
A bit complicated.
Are you saying that ##\tan \theta = X## has two answers?
That one of them is ##\arctan X## and the other one is ##\arctan X + \pi ## ...
Did I understand your idea correctly?
 
  • #13
MatinSAR said:
Now it is clear. Thanks for your help @BvU ...
I was using google calculator and it didn't gave exact value ... I will use wolframalpha.com from now.

A bit complicated.
Are you saying that ##\tan \theta = X## has two answers? And one of them is ##\arctan X## and the other one is ##\arctan X + \pi ## ...
It has infinitely-many, unless you restrict the range of values for ##X##, given ##arctanX ## is periodic.
 
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  • #14
MatinSAR said:
Are you saying that ##\tan \theta = X## has two answers? And one of them is ##\arctan X## and the other one is ##\arctan X + \pi ## ...
Yes. All three basic trig functions are 2:1 on the interval ##[0, 2\pi)##. In general:
$$\sin^{-1}(\sin x) \ne x \ \ (x \in [0, 2\pi))$$etc.
 
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  • #15
WWGD said:
It has infinitely-many, unless you restrict the range of values for ##X##, given ##arctanX ## is periodic.
PeroK said:
Yes. All three basic trig functions are 2:1 on the interval ##[0, 2\pi)##. In general:
$$\sin^{-1}(\sin x) \ne x$$etc.
I understand now. Thanks for your time.


Thank you to everyone for his help.
 
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