Dtjytyk

Extracting Dirichlet series coefficients

Magical attacks and overcoming damage resistance

What makes accurate emulation of old systems a difficult task?

Why did Rep. Omar conclude her criticism of US troops with the phrase "NotTodaySatan"?

Apply a different color ramp to subset of categorized symbols in QGIS?

How do I produce this Greek letter koppa: Ϟ in pdfLaTeX?

Find a stone which is not the lightest one

A strange hotel

How exactly does Hawking radiation decrease the mass of black holes?

Older movie/show about humans on derelict alien warship which refuels by passing through a star

All ASCII characters with a given bit count

Unknown code in script

Why did C use the -> operator instead of reusing the . operator?

SFDX - Create Objects with Custom Properties

Restricting the options of a lookup field, based on the value of another lookup field?

Mistake in years of experience in resume?

How to keep bees out of canned beverages?

How bug prioritization works in agile projects vs non agile

How to have a sharp product image?

What is purpose of DB Browser(dbbrowser.aspx) under admin tool?

Should the Product Owner dictate what info the UI needs to display?

Nails holding drywall

What does MLD stand for?

What is the best way to deal with NPC-NPC combat?

Complex numbers $z=-3-4i$ polar form

Expressing $e^z$ where $z=a+bi$ in polar form.Polar form of Complex numbersConverting to polar formComplex number polar form equationWhy is e used for polar form of complex numbers?Show $-27$ in polar form.writing in polar form (complex numbers)Are all complex exponentials in polar form?Complex numbers polar form changeDifferentiating polar functions using complex numbers

5

$begingroup$

Let $z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.

complex-numbers polar-coordinates

share|cite|improve this question

edited 10 mins ago

Thomas Shelby

4,8382727

asked 9 hours ago

user221435user221435

335

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice question! I wish more people thought about their questions before posting.
  $endgroup$
  – mackycheese21
  2 hours ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

Let $z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.

complex-numbers polar-coordinates

share|cite|improve this question

edited 10 mins ago

Thomas Shelby

4,8382727

asked 9 hours ago

user221435user221435

335

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice question! I wish more people thought about their questions before posting.
  $endgroup$
  – mackycheese21
  2 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.

complex-numbers polar-coordinates

share|cite|improve this question

edited 10 mins ago

Thomas Shelby

4,8382727

asked 9 hours ago

user221435user221435

335

$endgroup$

share|cite|improve this question

edited 10 mins ago

Thomas Shelby

4,8382727

asked 9 hours ago

user221435user221435

335

share|cite|improve this question

edited 10 mins ago

Thomas Shelby

4,8382727

asked 9 hours ago

user221435user221435

335

edited 10 mins ago

Thomas Shelby

4,8382727

edited 10 mins ago

Thomas Shelby

4,8382727

asked 9 hours ago

user221435user221435

335

asked 9 hours ago

user221435user221435

335

3 Answers
3

active

oldest

votes

6

$begingroup$

It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$

share|cite|improve this answer

answered 9 hours ago

José Carlos SantosJosé Carlos Santos

178k24139251

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
  $endgroup$
  – user221435
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
  $endgroup$
  – José Carlos Santos
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thank you
  $endgroup$
  – user221435
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
  $endgroup$
  – DreamConspiracy
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Being very fond of the geometrical plane of complex numbers, I feel that this is backwards (if not formally, then at least intuitively). In my opinion, $sin$ and $cos$ are unchanged after increasing or decreasing an angle by $360^circ$ because turning something $360^circ$ around the origin puts it back where you started. This sounds more like you're saying that turning something $360^circ$ around the origin gives you what you started because $sin$ and $cos$ are unchanged.
  $endgroup$
  – Arthur
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

3

$begingroup$

The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.

By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.

share|cite|improve this answer

answered 6 hours ago

ArthurArthur

123k7122211

$endgroup$

add a comment | 

2

$begingroup$

It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.

The principal angle is an angle between $-180^circ$ and $+180^circ$

share|cite|improve this answer

edited 8 hours ago

J. W. Tanner

5,2001520

answered 9 hours ago

Fareed AFFareed AF

1,025214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Acomment above shows the range [0∘,360). Curious why the difference.
  $endgroup$
  – JoeTaxpayer
  3 hours ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3202145%2fcomplex-numbers-z-3-4i-polar-form%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$

share|cite|improve this answer

answered 9 hours ago

José Carlos SantosJosé Carlos Santos

178k24139251

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
  $endgroup$
  – user221435
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
  $endgroup$
  – José Carlos Santos
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thank you
  $endgroup$
  – user221435
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
  $endgroup$
  – DreamConspiracy
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Being very fond of the geometrical plane of complex numbers, I feel that this is backwards (if not formally, then at least intuitively). In my opinion, $sin$ and $cos$ are unchanged after increasing or decreasing an angle by $360^circ$ because turning something $360^circ$ around the origin puts it back where you started. This sounds more like you're saying that turning something $360^circ$ around the origin gives you what you started because $sin$ and $cos$ are unchanged.
  $endgroup$
  – Arthur
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

6

$begingroup$

It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$

share|cite|improve this answer

answered 9 hours ago

José Carlos SantosJosé Carlos Santos

178k24139251

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
  $endgroup$
  – user221435
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
  $endgroup$
  – José Carlos Santos
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thank you
  $endgroup$
  – user221435
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
  $endgroup$
  – DreamConspiracy
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Being very fond of the geometrical plane of complex numbers, I feel that this is backwards (if not formally, then at least intuitively). In my opinion, $sin$ and $cos$ are unchanged after increasing or decreasing an angle by $360^circ$ because turning something $360^circ$ around the origin puts it back where you started. This sounds more like you're saying that turning something $360^circ$ around the origin gives you what you started because $sin$ and $cos$ are unchanged.
  $endgroup$
  – Arthur
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$

share|cite|improve this answer

answered 9 hours ago

José Carlos SantosJosé Carlos Santos

178k24139251

$endgroup$

share|cite|improve this answer

answered 9 hours ago

José Carlos SantosJosé Carlos Santos

178k24139251

answered 9 hours ago

José Carlos SantosJosé Carlos Santos

178k24139251

answered 9 hours ago

José Carlos SantosJosé Carlos Santos

178k24139251

3

$begingroup$

The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.

By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.

share|cite|improve this answer

answered 6 hours ago

ArthurArthur

123k7122211

$endgroup$

add a comment | 

3

$begingroup$

The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.

By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.

share|cite|improve this answer

answered 6 hours ago

ArthurArthur

123k7122211

$endgroup$

add a comment | 

$begingroup$

The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.

By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.

share|cite|improve this answer

answered 6 hours ago

ArthurArthur

123k7122211

$endgroup$

share|cite|improve this answer

answered 6 hours ago

ArthurArthur

123k7122211

answered 6 hours ago

ArthurArthur

123k7122211

answered 6 hours ago

ArthurArthur

123k7122211

2

$begingroup$

It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.

The principal angle is an angle between $-180^circ$ and $+180^circ$

share|cite|improve this answer

edited 8 hours ago

J. W. Tanner

5,2001520

answered 9 hours ago

Fareed AFFareed AF

1,025214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Acomment above shows the range [0∘,360). Curious why the difference.
  $endgroup$
  – JoeTaxpayer
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.

The principal angle is an angle between $-180^circ$ and $+180^circ$

share|cite|improve this answer

edited 8 hours ago

J. W. Tanner

5,2001520

answered 9 hours ago

Fareed AFFareed AF

1,025214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Acomment above shows the range [0∘,360). Curious why the difference.
  $endgroup$
  – JoeTaxpayer
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.

The principal angle is an angle between $-180^circ$ and $+180^circ$

share|cite|improve this answer

edited 8 hours ago

J. W. Tanner

5,2001520

answered 9 hours ago

Fareed AFFareed AF

1,025214

$endgroup$

share|cite|improve this answer

edited 8 hours ago

J. W. Tanner

5,2001520

answered 9 hours ago

Fareed AFFareed AF

1,025214

edited 8 hours ago

J. W. Tanner

5,2001520

edited 8 hours ago

J. W. Tanner

5,2001520

answered 9 hours ago

Fareed AFFareed AF

1,025214

answered 9 hours ago

Fareed AFFareed AF

1,025214