Why Is My Argument Calculation for Complex Number Incorrect?

[MatinSAR]

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.

 

[MatinSAR]

I've checked 2nd question using Microsoft Copilot. It was correct. I did not expect that it can calculate things like this ...

 

[BvU]

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

$\$

 

[FactChecker]

For another approach, which I think is simpler, try to relate $z=-\sin(\frac {\pi}{8}) -\cos(\frac {\pi}{8})$ to $e^{i\frac{\pi}{8}} = \cos(\frac {\pi}{8}) +i \sin(\frac {\pi}{8})$.

 

[pasmith]

For (1), note that $-z = \sin \frac\pi 8 + i\cos \frac \pi 8$.
Note also that $$\begin{split} \cos \theta &= \sin (\frac12\pi - \theta) \\ \sin \theta &= \cos(\frac12\pi - \theta).\end{split}$$ Hence $\sin \frac\pi 8 + i \cos \frac \pi 8 = \cos \frac {3 \pi} 8 + i \sin \frac {3\pi}8$.

 

[MatinSAR]

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

$\$

I've asked first question but the answer was unreadable since my android device doesn't support latex.

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.

For (1), note that $-z = \sin \frac\pi 8 + i\cos \frac \pi 8$.
Note also that $$\begin{split} \cos \theta &= \sin (\frac12\pi - \theta) \\ \sin \theta &= \cos(\frac12\pi - \theta).\end{split}$$ 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.

 

[PeroK]

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:

What I've done:
$$\tan \theta=\cot \frac {\pi}{8}$$

 

[WWGD]

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.

 

[MatinSAR]

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 $\tan \theta = \cot (\pi + \dfrac {\pi} {8})$?

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.

 

[BvU]

I've asked first question but the answer was unreadable since my android device doesn't support latex.

here's another crutch

$\$

 

[PeroK]

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.

 

[MatinSAR]

here's another crutch

$\$

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.

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?

 

[WWGD]

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.

 

[PeroK]

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.

 

[MatinSAR]

It has infinitely-many, unless you restrict the range of values for $X$, given $arctanX$ is periodic.

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.