Find $arg(z)$, where $z = -fracsqrt7(1+i)sqrt3+i$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












This is my work so far:



I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:



$-fracsqrt7(sqrt3+1)4 + fracsqrt7(1-sqrt3)4i$



When I solve for $|z|$ I get $fracsqrt142$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $frac11pi12$, while the book lists it as $frac13pi12$. I got my answer by solving for $cos (theta) = fraca$, where $theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.



Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!










share|cite|improve this question























  • And I get $pi/12$. I didn't bother with the tedious algebra though.
    – Lord Shark the Unknown
    Aug 30 at 3:48











  • I got the expression into the form a+bi That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
    – dxiv
    Aug 30 at 3:50











  • How'd you come to that answer without the algebra?
    – Leorio Paradinight
    Aug 30 at 3:50














up vote
0
down vote

favorite












This is my work so far:



I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:



$-fracsqrt7(sqrt3+1)4 + fracsqrt7(1-sqrt3)4i$



When I solve for $|z|$ I get $fracsqrt142$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $frac11pi12$, while the book lists it as $frac13pi12$. I got my answer by solving for $cos (theta) = fraca$, where $theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.



Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!










share|cite|improve this question























  • And I get $pi/12$. I didn't bother with the tedious algebra though.
    – Lord Shark the Unknown
    Aug 30 at 3:48











  • I got the expression into the form a+bi That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
    – dxiv
    Aug 30 at 3:50











  • How'd you come to that answer without the algebra?
    – Leorio Paradinight
    Aug 30 at 3:50












up vote
0
down vote

favorite









up vote
0
down vote

favorite











This is my work so far:



I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:



$-fracsqrt7(sqrt3+1)4 + fracsqrt7(1-sqrt3)4i$



When I solve for $|z|$ I get $fracsqrt142$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $frac11pi12$, while the book lists it as $frac13pi12$. I got my answer by solving for $cos (theta) = fraca$, where $theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.



Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!










share|cite|improve this question















This is my work so far:



I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:



$-fracsqrt7(sqrt3+1)4 + fracsqrt7(1-sqrt3)4i$



When I solve for $|z|$ I get $fracsqrt142$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $frac11pi12$, while the book lists it as $frac13pi12$. I got my answer by solving for $cos (theta) = fraca$, where $theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.



Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!







complex-analysis complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 30 at 3:56

























asked Aug 30 at 3:47









Leorio Paradinight

112




112











  • And I get $pi/12$. I didn't bother with the tedious algebra though.
    – Lord Shark the Unknown
    Aug 30 at 3:48











  • I got the expression into the form a+bi That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
    – dxiv
    Aug 30 at 3:50











  • How'd you come to that answer without the algebra?
    – Leorio Paradinight
    Aug 30 at 3:50
















  • And I get $pi/12$. I didn't bother with the tedious algebra though.
    – Lord Shark the Unknown
    Aug 30 at 3:48











  • I got the expression into the form a+bi That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
    – dxiv
    Aug 30 at 3:50











  • How'd you come to that answer without the algebra?
    – Leorio Paradinight
    Aug 30 at 3:50















And I get $pi/12$. I didn't bother with the tedious algebra though.
– Lord Shark the Unknown
Aug 30 at 3:48





And I get $pi/12$. I didn't bother with the tedious algebra though.
– Lord Shark the Unknown
Aug 30 at 3:48













I got the expression into the form a+bi That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
– dxiv
Aug 30 at 3:50





I got the expression into the form a+bi That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
– dxiv
Aug 30 at 3:50













How'd you come to that answer without the algebra?
– Leorio Paradinight
Aug 30 at 3:50




How'd you come to that answer without the algebra?
– Leorio Paradinight
Aug 30 at 3:50










1 Answer
1






active

oldest

votes

















up vote
2
down vote













$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$






share|cite|improve this answer




















  • It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
    – Leorio Paradinight
    Aug 30 at 3:54










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


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2899093%2ffind-argz-where-z-frac-sqrt71i-sqrt3i%23new-answer', 'question_page');

);

Post as a guest






























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$






share|cite|improve this answer




















  • It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
    – Leorio Paradinight
    Aug 30 at 3:54














up vote
2
down vote













$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$






share|cite|improve this answer




















  • It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
    – Leorio Paradinight
    Aug 30 at 3:54












up vote
2
down vote










up vote
2
down vote









$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$






share|cite|improve this answer












$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 30 at 3:51









Hello_World

3,20321429




3,20321429











  • It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
    – Leorio Paradinight
    Aug 30 at 3:54
















  • It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
    – Leorio Paradinight
    Aug 30 at 3:54















It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
– Leorio Paradinight
Aug 30 at 3:54




It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
– Leorio Paradinight
Aug 30 at 3:54

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2899093%2ffind-argz-where-z-frac-sqrt71i-sqrt3i%23new-answer', 'question_page');

);

Post as a guest













































































這個網誌中的熱門文章

How to combine Bézier curves to a surface?

Mutual Information Always Non-negative

Why am i infinitely getting the same tweet with the Twitter Search API?