Evaluate $(1+i)^(1-2i)$

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

1

Find all the values of $(1+i)^(1-2i)$ and show that there are small values as we wish (else from $0$) and big values as we wish

beginalign
(1+i)^(1-2i)&=e^ln(1+i)^(1-2i)
=e^(1-2i)ln(1+i)
\&=e^(lnsqrt2+i(fracpi2+2pi k))(1-2i)
\&=e^lnsqrt2-2lnsqrt2*i+i(fracpi2+2pi k)+2(fracpi2+2pi k))
\&=sqrt2*e^-2lnsqrt2*i*i*e^2(fracpi2+2pi k)
\&=e^-2lnsqrt2*i+pi+4pi k+isqrt2
\&=e^-i(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=e^-i*e^(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=frac34*e^(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=frac34*2e^i(pi+4pi k)+isqrt2
\&=frac-32+isqrt2
endalign

What can I conclude from it on the values?

Or should I stop here $e^-2lnsqrt2*i+pi+4pi k+isqrt2$ and $e^pi+4 pi kcis(-2lnsqrt2+fracpi2)$

complex-analysis

share|cite|improve this question

edited Jul 17 at 7:59

LutzL

49.8k31849

asked Jul 17 at 7:36

gbox

5,30851841

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is quite difficult to read. Consider using beginalign and endalign.
  – TheSimpliFire
  Jul 17 at 7:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TheSimpliFire added
  – gbox
  Jul 17 at 7:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In line 7 you used the wrong formula $e^-iz=e^-ie^z$.
  – Kavi Rama Murthy
  Jul 17 at 7:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somehow along the way (line 4->5) one multiplication mutated into an addition. You should not get a sum from your formula.
  – LutzL
  Jul 17 at 8:00
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

Find all the values of $(1+i)^(1-2i)$ and show that there are small values as we wish (else from $0$) and big values as we wish

beginalign
(1+i)^(1-2i)&=e^ln(1+i)^(1-2i)
=e^(1-2i)ln(1+i)
\&=e^(lnsqrt2+i(fracpi2+2pi k))(1-2i)
\&=e^lnsqrt2-2lnsqrt2*i+i(fracpi2+2pi k)+2(fracpi2+2pi k))
\&=sqrt2*e^-2lnsqrt2*i*i*e^2(fracpi2+2pi k)
\&=e^-2lnsqrt2*i+pi+4pi k+isqrt2
\&=e^-i(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=e^-i*e^(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=frac34*e^(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=frac34*2e^i(pi+4pi k)+isqrt2
\&=frac-32+isqrt2
endalign

What can I conclude from it on the values?

Or should I stop here $e^-2lnsqrt2*i+pi+4pi k+isqrt2$ and $e^pi+4 pi kcis(-2lnsqrt2+fracpi2)$

complex-analysis

share|cite|improve this question

edited Jul 17 at 7:59

LutzL

49.8k31849

asked Jul 17 at 7:36

gbox

5,30851841

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is quite difficult to read. Consider using beginalign and endalign.
  – TheSimpliFire
  Jul 17 at 7:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TheSimpliFire added
  – gbox
  Jul 17 at 7:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In line 7 you used the wrong formula $e^-iz=e^-ie^z$.
  – Kavi Rama Murthy
  Jul 17 at 7:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somehow along the way (line 4->5) one multiplication mutated into an addition. You should not get a sum from your formula.
  – LutzL
  Jul 17 at 8:00
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Find all the values of $(1+i)^(1-2i)$ and show that there are small values as we wish (else from $0$) and big values as we wish

beginalign
(1+i)^(1-2i)&=e^ln(1+i)^(1-2i)
=e^(1-2i)ln(1+i)
\&=e^(lnsqrt2+i(fracpi2+2pi k))(1-2i)
\&=e^lnsqrt2-2lnsqrt2*i+i(fracpi2+2pi k)+2(fracpi2+2pi k))
\&=sqrt2*e^-2lnsqrt2*i*i*e^2(fracpi2+2pi k)
\&=e^-2lnsqrt2*i+pi+4pi k+isqrt2
\&=e^-i(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=e^-i*e^(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=frac34*e^(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=frac34*2e^i(pi+4pi k)+isqrt2
\&=frac-32+isqrt2
endalign

What can I conclude from it on the values?

Or should I stop here $e^-2lnsqrt2*i+pi+4pi k+isqrt2$ and $e^pi+4 pi kcis(-2lnsqrt2+fracpi2)$

complex-analysis

share|cite|improve this question

edited Jul 17 at 7:59

LutzL

49.8k31849

asked Jul 17 at 7:36

gbox

5,30851841

Find all the values of $(1+i)^(1-2i)$ and show that there are small values as we wish (else from $0$) and big values as we wish

beginalign
(1+i)^(1-2i)&=e^ln(1+i)^(1-2i)
=e^(1-2i)ln(1+i)
\&=e^(lnsqrt2+i(fracpi2+2pi k))(1-2i)
\&=e^lnsqrt2-2lnsqrt2*i+i(fracpi2+2pi k)+2(fracpi2+2pi k))
\&=sqrt2*e^-2lnsqrt2*i*i*e^2(fracpi2+2pi k)
\&=e^-2lnsqrt2*i+pi+4pi k+isqrt2
\&=e^-i(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=e^-i*e^(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=frac34*e^(2lnsqrt2+i(pi+4pi k))+isqrt2
\&=frac34*2e^i(pi+4pi k)+isqrt2
\&=frac-32+isqrt2
endalign

What can I conclude from it on the values?

Or should I stop here $e^-2lnsqrt2*i+pi+4pi k+isqrt2$ and $e^pi+4 pi kcis(-2lnsqrt2+fracpi2)$

complex-analysis

share|cite|improve this question

edited Jul 17 at 7:59

LutzL

49.8k31849

asked Jul 17 at 7:36

gbox

5,30851841

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 7:59

LutzL

49.8k31849

edited Jul 17 at 7:59

LutzL

49.8k31849

edited Jul 17 at 7:59

LutzL

49.8k31849

49.8k31849

asked Jul 17 at 7:36

gbox

5,30851841

asked Jul 17 at 7:36

gbox

5,30851841

asked Jul 17 at 7:36

gbox

5,30851841

5,30851841

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is quite difficult to read. Consider using beginalign and endalign.
  – TheSimpliFire
  Jul 17 at 7:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TheSimpliFire added
  – gbox
  Jul 17 at 7:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In line 7 you used the wrong formula $e^-iz=e^-ie^z$.
  – Kavi Rama Murthy
  Jul 17 at 7:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somehow along the way (line 4->5) one multiplication mutated into an addition. You should not get a sum from your formula.
  – LutzL
  Jul 17 at 8:00
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is quite difficult to read. Consider using beginalign and endalign.
  – TheSimpliFire
  Jul 17 at 7:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TheSimpliFire added
  – gbox
  Jul 17 at 7:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In line 7 you used the wrong formula $e^-iz=e^-ie^z$.
  – Kavi Rama Murthy
  Jul 17 at 7:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somehow along the way (line 4->5) one multiplication mutated into an addition. You should not get a sum from your formula.
  – LutzL
  Jul 17 at 8:00
  
  

  
  
  
  

1

1

This is quite difficult to read. Consider using beginalign and endalign.
– TheSimpliFire
Jul 17 at 7:46

This is quite difficult to read. Consider using beginalign and endalign.
– TheSimpliFire
Jul 17 at 7:46

@TheSimpliFire added
– gbox
Jul 17 at 7:52

@TheSimpliFire added
– gbox
Jul 17 at 7:52

2

2

In line 7 you used the wrong formula $e^-iz=e^-ie^z$.
– Kavi Rama Murthy
Jul 17 at 7:59

In line 7 you used the wrong formula $e^-iz=e^-ie^z$.
– Kavi Rama Murthy
Jul 17 at 7:59

Somehow along the way (line 4->5) one multiplication mutated into an addition. You should not get a sum from your formula.
– LutzL
Jul 17 at 8:00

Somehow along the way (line 4->5) one multiplication mutated into an addition. You should not get a sum from your formula.
– LutzL
Jul 17 at 8:00

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

Hint. Note that $|e^z|=|e^textRe(z)+itextIm(z)|=e^textRe(z)$, hence here we are interested in (see your third step where the correct argument of $(1+i)$ is $pi/4+2pi k$)
$$textRe(lnsqrt2+i(fracpi4+2pi k))(1-2i))=lnsqrt2+pileft(4k+frac12right)$$
where $kinmathbbZ$. Now consider the limits as $kto pm infty$.

share|cite|improve this answer

edited Jul 17 at 8:30

answered Jul 17 at 7:59

Robert Z

84.2k955123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should it be $lnsqrt2+pi (4k+frac12)$?
  – gbox
  Jul 17 at 8:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you are right.
  – Robert Z
  Jul 17 at 8:30
  

  
  
  
  

add a comment | 

up vote
0
down vote

There should only be one value.$$(1+i)^1-2i=(sqrt 2e^dfracipi4)^1-2i=sqrt 2e^dfracipi4(0.5e^dfrac-ipi2)^i=sqrt 2e^dfracpi2e^i(dfracpi4+ln 0.5)$$

share|cite|improve this answer

answered Jul 17 at 8:04

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is not. $ $
  – Did
  Jul 17 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What you said means that $(dfrac12)^i$ should have different values. Why should it be so?
  – Mostafa Ayaz
  Jul 17 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should be the one telling us why -- or why you choose to answer questions on subjects you obviously do not master?
  – Did
  Jul 17 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wooow not so fast! If one quantity or expression is multivalued one must say why is that so not the one ascribing only one variable to it, for example nobody asks why $2^2$ has only one value but if somebody claims another value he should explain.
  – Mostafa Ayaz
  Jul 17 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, "so fast". You may not be realizing it but, posting this, you are squarely ignoring the basics of the domain (call it "complex analysis" if you wish). Again, you should explain why you see fit to post wrong answers.
  – Did
  Jul 17 at 8:55
  

  
  
  
  

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: "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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854264%2fevaluate-1i1-2i%23new-answer', 'question_page');

);

Post as a guest

Name Email

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Hint. Note that $|e^z|=|e^textRe(z)+itextIm(z)|=e^textRe(z)$, hence here we are interested in (see your third step where the correct argument of $(1+i)$ is $pi/4+2pi k$)
$$textRe(lnsqrt2+i(fracpi4+2pi k))(1-2i))=lnsqrt2+pileft(4k+frac12right)$$
where $kinmathbbZ$. Now consider the limits as $kto pm infty$.

share|cite|improve this answer

edited Jul 17 at 8:30

answered Jul 17 at 7:59

Robert Z

84.2k955123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should it be $lnsqrt2+pi (4k+frac12)$?
  – gbox
  Jul 17 at 8:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you are right.
  – Robert Z
  Jul 17 at 8:30
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Hint. Note that $|e^z|=|e^textRe(z)+itextIm(z)|=e^textRe(z)$, hence here we are interested in (see your third step where the correct argument of $(1+i)$ is $pi/4+2pi k$)
$$textRe(lnsqrt2+i(fracpi4+2pi k))(1-2i))=lnsqrt2+pileft(4k+frac12right)$$
where $kinmathbbZ$. Now consider the limits as $kto pm infty$.

share|cite|improve this answer

edited Jul 17 at 8:30

answered Jul 17 at 7:59

Robert Z

84.2k955123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should it be $lnsqrt2+pi (4k+frac12)$?
  – gbox
  Jul 17 at 8:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you are right.
  – Robert Z
  Jul 17 at 8:30
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Hint. Note that $|e^z|=|e^textRe(z)+itextIm(z)|=e^textRe(z)$, hence here we are interested in (see your third step where the correct argument of $(1+i)$ is $pi/4+2pi k$)
$$textRe(lnsqrt2+i(fracpi4+2pi k))(1-2i))=lnsqrt2+pileft(4k+frac12right)$$
where $kinmathbbZ$. Now consider the limits as $kto pm infty$.

share|cite|improve this answer

edited Jul 17 at 8:30

answered Jul 17 at 7:59

Robert Z

84.2k955123

Hint. Note that $|e^z|=|e^textRe(z)+itextIm(z)|=e^textRe(z)$, hence here we are interested in (see your third step where the correct argument of $(1+i)$ is $pi/4+2pi k$)
$$textRe(lnsqrt2+i(fracpi4+2pi k))(1-2i))=lnsqrt2+pileft(4k+frac12right)$$
where $kinmathbbZ$. Now consider the limits as $kto pm infty$.

share|cite|improve this answer

edited Jul 17 at 8:30

answered Jul 17 at 7:59

Robert Z

84.2k955123

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 8:30

edited Jul 17 at 8:30

edited Jul 17 at 8:30

answered Jul 17 at 7:59

Robert Z

84.2k955123

answered Jul 17 at 7:59

Robert Z

84.2k955123

answered Jul 17 at 7:59

Robert Z

84.2k955123

84.2k955123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should it be $lnsqrt2+pi (4k+frac12)$?
  – gbox
  Jul 17 at 8:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you are right.
  – Robert Z
  Jul 17 at 8:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should it be $lnsqrt2+pi (4k+frac12)$?
  – gbox
  Jul 17 at 8:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you are right.
  – Robert Z
  Jul 17 at 8:30
  

  
  
  
  

Should it be $lnsqrt2+pi (4k+frac12)$?
– gbox
Jul 17 at 8:15

Should it be $lnsqrt2+pi (4k+frac12)$?
– gbox
Jul 17 at 8:15

Yes, you are right.
– Robert Z
Jul 17 at 8:30

Yes, you are right.
– Robert Z
Jul 17 at 8:30

add a comment | 

up vote
0
down vote

There should only be one value.$$(1+i)^1-2i=(sqrt 2e^dfracipi4)^1-2i=sqrt 2e^dfracipi4(0.5e^dfrac-ipi2)^i=sqrt 2e^dfracpi2e^i(dfracpi4+ln 0.5)$$

share|cite|improve this answer

answered Jul 17 at 8:04

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is not. $ $
  – Did
  Jul 17 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What you said means that $(dfrac12)^i$ should have different values. Why should it be so?
  – Mostafa Ayaz
  Jul 17 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should be the one telling us why -- or why you choose to answer questions on subjects you obviously do not master?
  – Did
  Jul 17 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wooow not so fast! If one quantity or expression is multivalued one must say why is that so not the one ascribing only one variable to it, for example nobody asks why $2^2$ has only one value but if somebody claims another value he should explain.
  – Mostafa Ayaz
  Jul 17 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, "so fast". You may not be realizing it but, posting this, you are squarely ignoring the basics of the domain (call it "complex analysis" if you wish). Again, you should explain why you see fit to post wrong answers.
  – Did
  Jul 17 at 8:55
  

  
  
  
  

add a comment | 

up vote
0
down vote

There should only be one value.$$(1+i)^1-2i=(sqrt 2e^dfracipi4)^1-2i=sqrt 2e^dfracipi4(0.5e^dfrac-ipi2)^i=sqrt 2e^dfracpi2e^i(dfracpi4+ln 0.5)$$

share|cite|improve this answer

answered Jul 17 at 8:04

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is not. $ $
  – Did
  Jul 17 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What you said means that $(dfrac12)^i$ should have different values. Why should it be so?
  – Mostafa Ayaz
  Jul 17 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should be the one telling us why -- or why you choose to answer questions on subjects you obviously do not master?
  – Did
  Jul 17 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wooow not so fast! If one quantity or expression is multivalued one must say why is that so not the one ascribing only one variable to it, for example nobody asks why $2^2$ has only one value but if somebody claims another value he should explain.
  – Mostafa Ayaz
  Jul 17 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, "so fast". You may not be realizing it but, posting this, you are squarely ignoring the basics of the domain (call it "complex analysis" if you wish). Again, you should explain why you see fit to post wrong answers.
  – Did
  Jul 17 at 8:55
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

There should only be one value.$$(1+i)^1-2i=(sqrt 2e^dfracipi4)^1-2i=sqrt 2e^dfracipi4(0.5e^dfrac-ipi2)^i=sqrt 2e^dfracpi2e^i(dfracpi4+ln 0.5)$$

share|cite|improve this answer

answered Jul 17 at 8:04

Mostafa Ayaz

8,6023630

There should only be one value.$$(1+i)^1-2i=(sqrt 2e^dfracipi4)^1-2i=sqrt 2e^dfracipi4(0.5e^dfrac-ipi2)^i=sqrt 2e^dfracpi2e^i(dfracpi4+ln 0.5)$$

share|cite|improve this answer

answered Jul 17 at 8:04

Mostafa Ayaz

8,6023630

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 8:04

Mostafa Ayaz

8,6023630

answered Jul 17 at 8:04

Mostafa Ayaz

8,6023630

answered Jul 17 at 8:04

Mostafa Ayaz

8,6023630

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is not. $ $
  – Did
  Jul 17 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What you said means that $(dfrac12)^i$ should have different values. Why should it be so?
  – Mostafa Ayaz
  Jul 17 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should be the one telling us why -- or why you choose to answer questions on subjects you obviously do not master?
  – Did
  Jul 17 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wooow not so fast! If one quantity or expression is multivalued one must say why is that so not the one ascribing only one variable to it, for example nobody asks why $2^2$ has only one value but if somebody claims another value he should explain.
  – Mostafa Ayaz
  Jul 17 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, "so fast". You may not be realizing it but, posting this, you are squarely ignoring the basics of the domain (call it "complex analysis" if you wish). Again, you should explain why you see fit to post wrong answers.
  – Did
  Jul 17 at 8:55
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is not. $ $
  – Did
  Jul 17 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What you said means that $(dfrac12)^i$ should have different values. Why should it be so?
  – Mostafa Ayaz
  Jul 17 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should be the one telling us why -- or why you choose to answer questions on subjects you obviously do not master?
  – Did
  Jul 17 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wooow not so fast! If one quantity or expression is multivalued one must say why is that so not the one ascribing only one variable to it, for example nobody asks why $2^2$ has only one value but if somebody claims another value he should explain.
  – Mostafa Ayaz
  Jul 17 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, "so fast". You may not be realizing it but, posting this, you are squarely ignoring the basics of the domain (call it "complex analysis" if you wish). Again, you should explain why you see fit to post wrong answers.
  – Did
  Jul 17 at 8:55
  

  
  
  
  

There is not. $ $
– Did
Jul 17 at 8:38

There is not. $ $
– Did
Jul 17 at 8:38

What you said means that $(dfrac12)^i$ should have different values. Why should it be so?
– Mostafa Ayaz
Jul 17 at 8:40

What you said means that $(dfrac12)^i$ should have different values. Why should it be so?
– Mostafa Ayaz
Jul 17 at 8:40

You should be the one telling us why -- or why you choose to answer questions on subjects you obviously do not master?
– Did
Jul 17 at 8:42

You should be the one telling us why -- or why you choose to answer questions on subjects you obviously do not master?
– Did
Jul 17 at 8:42

Wooow not so fast! If one quantity or expression is multivalued one must say why is that so not the one ascribing only one variable to it, for example nobody asks why $2^2$ has only one value but if somebody claims another value he should explain.
– Mostafa Ayaz
Jul 17 at 8:50

Wooow not so fast! If one quantity or expression is multivalued one must say why is that so not the one ascribing only one variable to it, for example nobody asks why $2^2$ has only one value but if somebody claims another value he should explain.
– Mostafa Ayaz
Jul 17 at 8:50

Yes, "so fast". You may not be realizing it but, posting this, you are squarely ignoring the basics of the domain (call it "complex analysis" if you wish). Again, you should explain why you see fit to post wrong answers.
– Did
Jul 17 at 8:55

Yes, "so fast". You may not be realizing it but, posting this, you are squarely ignoring the basics of the domain (call it "complex analysis" if you wish). Again, you should explain why you see fit to post wrong answers.
– Did
Jul 17 at 8:55

add a comment | 

 

draft saved

draft discarded

 

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%2fmath.stackexchange.com%2fquestions%2f2854264%2fevaluate-1i1-2i%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