i^π to trigonometric form

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

How can I convert the complex number $ mathrm i^pi $ to trigonometric form?

I usually do these steps:

1. take $ Z = a + bmathrm i $ form,


2. find $ r = sqrta^2 + b^2 $,


3. 
   $ cos(phi) = a / r, sin(phi) = b / r $,


4. find $ phi $ from the above 2 equations.

For $ mathrm i^pi $ I have $ a = 1, b = 1, r = 1 $, $ cos(phi) = 1, sin(phi) = 1 $.
There's no such $ phi $.

The online Convert Complex Numbers to Polar Form gives the answer $ phi = 77.2567 $ or just, $ phi = dfrac180 arg(mathrm i^pi)pi$

complex trigonometry

share|improve this question

edited 6 mins ago

Michael E2

142k11192460

asked 6 hours ago

user3132457

111

New contributor

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
2
down vote

favorite

How can I convert the complex number $ mathrm i^pi $ to trigonometric form?

I usually do these steps:

1. take $ Z = a + bmathrm i $ form,


2. find $ r = sqrta^2 + b^2 $,


3. 
   $ cos(phi) = a / r, sin(phi) = b / r $,


4. find $ phi $ from the above 2 equations.

For $ mathrm i^pi $ I have $ a = 1, b = 1, r = 1 $, $ cos(phi) = 1, sin(phi) = 1 $.
There's no such $ phi $.

The online Convert Complex Numbers to Polar Form gives the answer $ phi = 77.2567 $ or just, $ phi = dfrac180 arg(mathrm i^pi)pi$

complex trigonometry

share|improve this question

edited 6 mins ago

Michael E2

142k11192460

asked 6 hours ago

user3132457

111

New contributor

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

How can I convert the complex number $ mathrm i^pi $ to trigonometric form?

I usually do these steps:

1. take $ Z = a + bmathrm i $ form,


2. find $ r = sqrta^2 + b^2 $,


3. 
   $ cos(phi) = a / r, sin(phi) = b / r $,


4. find $ phi $ from the above 2 equations.

For $ mathrm i^pi $ I have $ a = 1, b = 1, r = 1 $, $ cos(phi) = 1, sin(phi) = 1 $.
There's no such $ phi $.

The online Convert Complex Numbers to Polar Form gives the answer $ phi = 77.2567 $ or just, $ phi = dfrac180 arg(mathrm i^pi)pi$

complex trigonometry

share|improve this question

edited 6 mins ago

Michael E2

142k11192460

asked 6 hours ago

user3132457

111

New contributor

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

How can I convert the complex number $ mathrm i^pi $ to trigonometric form?

I usually do these steps:

1. take $ Z = a + bmathrm i $ form,


2. find $ r = sqrta^2 + b^2 $,


3. 
   $ cos(phi) = a / r, sin(phi) = b / r $,


4. find $ phi $ from the above 2 equations.

For $ mathrm i^pi $ I have $ a = 1, b = 1, r = 1 $, $ cos(phi) = 1, sin(phi) = 1 $.
There's no such $ phi $.

The online Convert Complex Numbers to Polar Form gives the answer $ phi = 77.2567 $ or just, $ phi = dfrac180 arg(mathrm i^pi)pi$

complex trigonometry

complex trigonometry

share|improve this question

edited 6 mins ago

Michael E2

142k11192460

asked 6 hours ago

user3132457

111

New contributor

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 6 mins ago

Michael E2

142k11192460

asked 6 hours ago

user3132457

111

New contributor

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

share|improve this question

edited 6 mins ago

Michael E2

142k11192460

edited 6 mins ago

Michael E2

142k11192460

edited 6 mins ago

Michael E2

142k11192460

142k11192460

asked 6 hours ago

user3132457

111

New contributor

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 6 hours ago

user3132457

111

asked 6 hours ago

user3132457

111

111

New contributor

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

user3132457 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
6
down vote

ComplexExpand[I^π]

Cos[π^2/2] + I Sin[π^2/2]

share|improve this answer

answered 6 hours ago

That Gravity Guy

2,040514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i don't get the logic behind the calculations. where is the pi^2/2 from?
  – user3132457
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user3132457 $ mathrm i = mathrm e^mathrm i pi/2 $
  – Î‘λέξανδρος Ζεγγ
  4 hours ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

That Gravity Guy answers it very nicely, but if you want to see a drawn out calculation, first get your value into Exp[x] form:

I^Pi == E^x

Log[%[[1]]] == Log[%[[2]]] // PowerExpand
(*(I π^2)/2 == x*)

Solve[%, x] // Flatten
(*x -> (I π^2)/2*)

Now we can just convert to trig

ExpToTrig[E^x] /. %
(*Cos[π^2/2] + I Sin[π^2/2]*)

Get r, ϕ from

Re[%], Im[%] // ToPolarCoordinates // N
(*1., -1.34838*)

If you like degrees better

%[[1]], %[[2]]/°
(*1., -77.2567*)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Bill Watts

2,1781514

add a comment | 

up vote
0
down vote

So, first of all, you made a small mistake :
In your method, you have $a = 0$, and not $a=1$
that makes $b=1$, $r=1$, $cos(phi)=0$ and $sin(phi)=1$, that implies $phi = fracpi2 + 2kpi$ for any $k in mathbbZ$.

The problem is all values of $k$ are correct (as $cos(alpha) = cos(alpha+2pi)$, whatever the value of $alpha$, and we have the same for $sin$), and as such, $i^pi$ doesn't make any mathematical sense, since $e^ifracpi^22 neq e^ifrac5pi^22$, and mathematically speaking, you have no reason to chose one over the other.

Math software usually ignore this by arbitrarily choosing the argument value that is between $-pi$ and $pi$ (or sometimes $0$ and $2pi$, which even there can lead to different results if you took, for instance, "$(-i)^pi$" )

share|improve this answer

answered 1 hour ago

Aniem

1

New contributor

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
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',
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

user3132457 is a new contributor. Be nice, and check out our Code of Conduct.

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f185400%2fi%25cf%2580-to-trigonometric-form%23new-answer', 'question_page');

);

Post as a guest

Name Email

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote

ComplexExpand[I^π]

Cos[π^2/2] + I Sin[π^2/2]

share|improve this answer

answered 6 hours ago

That Gravity Guy

2,040514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i don't get the logic behind the calculations. where is the pi^2/2 from?
  – user3132457
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user3132457 $ mathrm i = mathrm e^mathrm i pi/2 $
  – Î‘λέξανδρος Ζεγγ
  4 hours ago
  

  
  
  
  

add a comment | 

up vote
6
down vote

ComplexExpand[I^π]

Cos[π^2/2] + I Sin[π^2/2]

share|improve this answer

answered 6 hours ago

That Gravity Guy

2,040514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i don't get the logic behind the calculations. where is the pi^2/2 from?
  – user3132457
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user3132457 $ mathrm i = mathrm e^mathrm i pi/2 $
  – Î‘λέξανδρος Ζεγγ
  4 hours ago
  

  
  
  
  

add a comment | 

up vote
6
down vote

up vote
6
down vote

ComplexExpand[I^π]

Cos[π^2/2] + I Sin[π^2/2]

share|improve this answer

answered 6 hours ago

That Gravity Guy

2,040514

ComplexExpand[I^π]

Cos[π^2/2] + I Sin[π^2/2]

share|improve this answer

answered 6 hours ago

That Gravity Guy

2,040514

share|improve this answer

share|improve this answer

answered 6 hours ago

That Gravity Guy

2,040514

answered 6 hours ago

That Gravity Guy

2,040514

answered 6 hours ago

That Gravity Guy

2,040514

2,040514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i don't get the logic behind the calculations. where is the pi^2/2 from?
  – user3132457
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user3132457 $ mathrm i = mathrm e^mathrm i pi/2 $
  – Î‘λέξανδρος Ζεγγ
  4 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i don't get the logic behind the calculations. where is the pi^2/2 from?
  – user3132457
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user3132457 $ mathrm i = mathrm e^mathrm i pi/2 $
  – Î‘λέξανδρος Ζεγγ
  4 hours ago
  

  
  
  
  

i don't get the logic behind the calculations. where is the pi^2/2 from?
– user3132457
5 hours ago

i don't get the logic behind the calculations. where is the pi^2/2 from?
– user3132457
5 hours ago

1

1

@user3132457 $ mathrm i = mathrm e^mathrm i pi/2 $
– Î‘λέξανδρος Ζεγγ
4 hours ago

@user3132457 $ mathrm i = mathrm e^mathrm i pi/2 $
– Î‘λέξανδρος Ζεγγ
4 hours ago

add a comment | 

up vote
4
down vote

That Gravity Guy answers it very nicely, but if you want to see a drawn out calculation, first get your value into Exp[x] form:

I^Pi == E^x

Log[%[[1]]] == Log[%[[2]]] // PowerExpand
(*(I π^2)/2 == x*)

Solve[%, x] // Flatten
(*x -> (I π^2)/2*)

Now we can just convert to trig

ExpToTrig[E^x] /. %
(*Cos[π^2/2] + I Sin[π^2/2]*)

Get r, ϕ from

Re[%], Im[%] // ToPolarCoordinates // N
(*1., -1.34838*)

If you like degrees better

%[[1]], %[[2]]/°
(*1., -77.2567*)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Bill Watts

2,1781514

add a comment | 

up vote
4
down vote

That Gravity Guy answers it very nicely, but if you want to see a drawn out calculation, first get your value into Exp[x] form:

I^Pi == E^x

Log[%[[1]]] == Log[%[[2]]] // PowerExpand
(*(I π^2)/2 == x*)

Solve[%, x] // Flatten
(*x -> (I π^2)/2*)

Now we can just convert to trig

ExpToTrig[E^x] /. %
(*Cos[π^2/2] + I Sin[π^2/2]*)

Get r, ϕ from

Re[%], Im[%] // ToPolarCoordinates // N
(*1., -1.34838*)

If you like degrees better

%[[1]], %[[2]]/°
(*1., -77.2567*)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Bill Watts

2,1781514

add a comment | 

up vote
4
down vote

up vote
4
down vote

That Gravity Guy answers it very nicely, but if you want to see a drawn out calculation, first get your value into Exp[x] form:

I^Pi == E^x

Log[%[[1]]] == Log[%[[2]]] // PowerExpand
(*(I π^2)/2 == x*)

Solve[%, x] // Flatten
(*x -> (I π^2)/2*)

Now we can just convert to trig

ExpToTrig[E^x] /. %
(*Cos[π^2/2] + I Sin[π^2/2]*)

Get r, ϕ from

Re[%], Im[%] // ToPolarCoordinates // N
(*1., -1.34838*)

If you like degrees better

%[[1]], %[[2]]/°
(*1., -77.2567*)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Bill Watts

2,1781514

That Gravity Guy answers it very nicely, but if you want to see a drawn out calculation, first get your value into Exp[x] form:

I^Pi == E^x

Log[%[[1]]] == Log[%[[2]]] // PowerExpand
(*(I π^2)/2 == x*)

Solve[%, x] // Flatten
(*x -> (I π^2)/2*)

Now we can just convert to trig

ExpToTrig[E^x] /. %
(*Cos[π^2/2] + I Sin[π^2/2]*)

Get r, ϕ from

Re[%], Im[%] // ToPolarCoordinates // N
(*1., -1.34838*)

If you like degrees better

%[[1]], %[[2]]/°
(*1., -77.2567*)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Bill Watts

2,1781514

share|improve this answer

share|improve this answer

edited 4 hours ago

edited 4 hours ago

edited 4 hours ago

answered 4 hours ago

Bill Watts

2,1781514

answered 4 hours ago

Bill Watts

2,1781514

answered 4 hours ago

Bill Watts

2,1781514

2,1781514

add a comment | 

add a comment | 

up vote
0
down vote

So, first of all, you made a small mistake :
In your method, you have $a = 0$, and not $a=1$
that makes $b=1$, $r=1$, $cos(phi)=0$ and $sin(phi)=1$, that implies $phi = fracpi2 + 2kpi$ for any $k in mathbbZ$.

The problem is all values of $k$ are correct (as $cos(alpha) = cos(alpha+2pi)$, whatever the value of $alpha$, and we have the same for $sin$), and as such, $i^pi$ doesn't make any mathematical sense, since $e^ifracpi^22 neq e^ifrac5pi^22$, and mathematically speaking, you have no reason to chose one over the other.

Math software usually ignore this by arbitrarily choosing the argument value that is between $-pi$ and $pi$ (or sometimes $0$ and $2pi$, which even there can lead to different results if you took, for instance, "$(-i)^pi$" )

share|improve this answer

answered 1 hour ago

Aniem

1

New contributor

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
0
down vote

So, first of all, you made a small mistake :
In your method, you have $a = 0$, and not $a=1$
that makes $b=1$, $r=1$, $cos(phi)=0$ and $sin(phi)=1$, that implies $phi = fracpi2 + 2kpi$ for any $k in mathbbZ$.

The problem is all values of $k$ are correct (as $cos(alpha) = cos(alpha+2pi)$, whatever the value of $alpha$, and we have the same for $sin$), and as such, $i^pi$ doesn't make any mathematical sense, since $e^ifracpi^22 neq e^ifrac5pi^22$, and mathematically speaking, you have no reason to chose one over the other.

Math software usually ignore this by arbitrarily choosing the argument value that is between $-pi$ and $pi$ (or sometimes $0$ and $2pi$, which even there can lead to different results if you took, for instance, "$(-i)^pi$" )

share|improve this answer

answered 1 hour ago

Aniem

1

New contributor

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
0
down vote

up vote
0
down vote

So, first of all, you made a small mistake :
In your method, you have $a = 0$, and not $a=1$
that makes $b=1$, $r=1$, $cos(phi)=0$ and $sin(phi)=1$, that implies $phi = fracpi2 + 2kpi$ for any $k in mathbbZ$.

The problem is all values of $k$ are correct (as $cos(alpha) = cos(alpha+2pi)$, whatever the value of $alpha$, and we have the same for $sin$), and as such, $i^pi$ doesn't make any mathematical sense, since $e^ifracpi^22 neq e^ifrac5pi^22$, and mathematically speaking, you have no reason to chose one over the other.

Math software usually ignore this by arbitrarily choosing the argument value that is between $-pi$ and $pi$ (or sometimes $0$ and $2pi$, which even there can lead to different results if you took, for instance, "$(-i)^pi$" )

share|improve this answer

answered 1 hour ago

Aniem

1

New contributor

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

So, first of all, you made a small mistake :
In your method, you have $a = 0$, and not $a=1$
that makes $b=1$, $r=1$, $cos(phi)=0$ and $sin(phi)=1$, that implies $phi = fracpi2 + 2kpi$ for any $k in mathbbZ$.

The problem is all values of $k$ are correct (as $cos(alpha) = cos(alpha+2pi)$, whatever the value of $alpha$, and we have the same for $sin$), and as such, $i^pi$ doesn't make any mathematical sense, since $e^ifracpi^22 neq e^ifrac5pi^22$, and mathematically speaking, you have no reason to chose one over the other.

Math software usually ignore this by arbitrarily choosing the argument value that is between $-pi$ and $pi$ (or sometimes $0$ and $2pi$, which even there can lead to different results if you took, for instance, "$(-i)^pi$" )

share|improve this answer

answered 1 hour ago

Aniem

1

New contributor

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this answer

share|improve this answer

answered 1 hour ago

Aniem

1

New contributor

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 1 hour ago

Aniem

1

answered 1 hour ago

Aniem

1

1

New contributor

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Aniem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

user3132457 is a new contributor. Be nice, and check out our Code of Conduct.

 

draft saved

draft discarded

user3132457 is a new contributor. Be nice, and check out our Code of Conduct.

user3132457 is a new contributor. Be nice, and check out our Code of Conduct.

user3132457 is a new contributor. Be nice, and check out our Code of Conduct.

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f185400%2fi%25cf%2580-to-trigonometric-form%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email