Why Is My Argument Calculation for Complex Number Incorrect?

AI Thread Summary
The discussion centers on calculating the argument of the complex number z = -sin(π/8) - i cos(π/8), with participants debating the correct value and its relationship to the unit circle. The modulus is confirmed to be 1, but there is confusion regarding the argument, with suggestions to relate z to exponential form and to consider the quadrant in which the number lies. Participants emphasize the importance of using the complex plane for clarity and note that the tangent function has multiple solutions, complicating the calculation of the argument. The second question about calculating (-1 + i√3)^{60} is confirmed correct, and there is a consensus on using reliable tools like Wolfram Alpha for complex calculations. Understanding the periodic nature of trigonometric functions is highlighted as crucial for accurate argument determination.
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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I've checked 2nd question using Microsoft Copilot. It was correct. I did not expect that it can calculate things like this ...
 
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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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})##.
 
For (1), note that -z = \sin \frac\pi 8 + i\cos \frac \pi 8.
Note also that \begin{split} \cos \theta &amp;= \sin (\frac12\pi - \theta) \\<br /> \sin \theta &amp;= \cos(\frac12\pi - \theta).\end{split} Hence \sin \frac\pi 8 + i \cos \frac \pi 8 = \cos \frac {3 \pi} 8 + i \sin \frac {3\pi}8.
 
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 -z = \sin \frac\pi 8 + i\cos \frac \pi 8.
Note also that \begin{split} \cos \theta &amp;= \sin (\frac12\pi - \theta) \\<br /> \sin \theta &amp;= \cos(\frac12\pi - \theta).\end{split} Hence \sin \frac\pi 8 + i \cos \frac \pi 8 = \cos \frac {3 \pi} 8 + i \sin \frac {3\pi}8.
Great! I've understand. Many thanks.
 
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:
$$\tan \theta=\cot \frac {\pi}{8}$$
 
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.
 
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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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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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.
 
  • #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.
 
  • #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.
 
  • #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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