Mixing
complex equation with a conjugate
$begingroup$
Find all the solutions to the equation:
$(1-i)z^{3}bar{z}=7+i$
My attempt:
$z^3bar{z}=frac{7+i}{1-i}=3+4i\$
$z^3bar{z}=3+4i rightarrow arg(z^3bar{z})=arg(3+4i)\$
$arg(z^3)-arg(z)=tan^{-1}(4/3)+2pi k\$
$2arg(z)=tan^{-1}(4/3)+2pi k\$
Therefore the angle is:
$arg(z)=frac{tan^{-1}(4/3)}{2}+pi k$
And to find the magnitude:
$| z^3bar{z}|=|z^3||z|=|z|^4=|3+4i| \$
$|z|^4=5 rightarrow |z|=sqrt[4]5\$
Therefore
$z=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)+ipi k}.$
By plugging $k = 0$ and $k = 1$ we get
$z_{1}=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)}=sqrt[4]5(cos(0.5tan^{-1}(4/3)) + isin(0.5tan^{-1}(4/3))$
$z_{2}=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)+ipi}\=sqrt[4]5(cos(0.5tan^{-1}(4/3)+pi) + isin(0.5tan^{-1}(4/3)+pi)$
Yet the correct solution is:
$z=pm sqrt[4]5(cos(0.5tan^{-1}(4/3)) + isin(0.5tan^{-1}(4/3))\$
So $z_1$ is a correct solution, but i'm not sure what about $z_2$ and why there is a negative solution in the answer.
Thanks
complex-numbers
edited Jan 20 at 19:47
J. W. Tanner
2,7911217
asked Jan 20 at 18:31
PulleyBoyPulleyBoy
31
$endgroup$
add a comment |
$begingroup$
Find all the solutions to the equation:
$(1-i)z^{3}bar{z}=7+i$
My attempt:
$z^3bar{z}=frac{7+i}{1-i}=3+4i\$
$z^3bar{z}=3+4i rightarrow arg(z^3bar{z})=arg(3+4i)\$
$arg(z^3)-arg(z)=tan^{-1}(4/3)+2pi k\$
$2arg(z)=tan^{-1}(4/3)+2pi k\$
Therefore the angle is:
$arg(z)=frac{tan^{-1}(4/3)}{2}+pi k$
And to find the magnitude:
$| z^3bar{z}|=|z^3||z|=|z|^4=|3+4i| \$
$|z|^4=5 rightarrow |z|=sqrt[4]5\$
Therefore
$z=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)+ipi k}.$
By plugging $k = 0$ and $k = 1$ we get
$z_{1}=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)}=sqrt[4]5(cos(0.5tan^{-1}(4/3)) + isin(0.5tan^{-1}(4/3))$
$z_{2}=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)+ipi}\=sqrt[4]5(cos(0.5tan^{-1}(4/3)+pi) + isin(0.5tan^{-1}(4/3)+pi)$
Yet the correct solution is:
$z=pm sqrt[4]5(cos(0.5tan^{-1}(4/3)) + isin(0.5tan^{-1}(4/3))\$
So $z_1$ is a correct solution, but i'm not sure what about $z_2$ and why there is a negative solution in the answer.
Thanks
complex-numbers
edited Jan 20 at 19:47
J. W. Tanner
2,7911217
asked Jan 20 at 18:31
PulleyBoyPulleyBoy
31
$endgroup$
add a comment |
$begingroup$
Find all the solutions to the equation:
$(1-i)z^{3}bar{z}=7+i$
My attempt:
$z^3bar{z}=frac{7+i}{1-i}=3+4i\$
$z^3bar{z}=3+4i rightarrow arg(z^3bar{z})=arg(3+4i)\$
$arg(z^3)-arg(z)=tan^{-1}(4/3)+2pi k\$
$2arg(z)=tan^{-1}(4/3)+2pi k\$
Therefore the angle is:
$arg(z)=frac{tan^{-1}(4/3)}{2}+pi k$
And to find the magnitude:
$| z^3bar{z}|=|z^3||z|=|z|^4=|3+4i| \$
$|z|^4=5 rightarrow |z|=sqrt[4]5\$
Therefore
$z=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)+ipi k}.$
By plugging $k = 0$ and $k = 1$ we get
$z_{1}=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)}=sqrt[4]5(cos(0.5tan^{-1}(4/3)) + isin(0.5tan^{-1}(4/3))$
$z_{2}=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)+ipi}\=sqrt[4]5(cos(0.5tan^{-1}(4/3)+pi) + isin(0.5tan^{-1}(4/3)+pi)$
Yet the correct solution is:
$z=pm sqrt[4]5(cos(0.5tan^{-1}(4/3)) + isin(0.5tan^{-1}(4/3))\$
So $z_1$ is a correct solution, but i'm not sure what about $z_2$ and why there is a negative solution in the answer.
Thanks
complex-numbers
edited Jan 20 at 19:47
J. W. Tanner
2,7911217
asked Jan 20 at 18:31
PulleyBoyPulleyBoy
31
$endgroup$
Find all the solutions to the equation:
$(1-i)z^{3}bar{z}=7+i$
My attempt:
$z^3bar{z}=frac{7+i}{1-i}=3+4i\$
$z^3bar{z}=3+4i rightarrow arg(z^3bar{z})=arg(3+4i)\$
$arg(z^3)-arg(z)=tan^{-1}(4/3)+2pi k\$
$2arg(z)=tan^{-1}(4/3)+2pi k\$
Therefore the angle is:
$arg(z)=frac{tan^{-1}(4/3)}{2}+pi k$
And to find the magnitude:
$| z^3bar{z}|=|z^3||z|=|z|^4=|3+4i| \$
$|z|^4=5 rightarrow |z|=sqrt[4]5\$
Therefore
$z=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)+ipi k}.$
By plugging $k = 0$ and $k = 1$ we get
$z_{1}=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)}=sqrt[4]5(cos(0.5tan^{-1}(4/3)) + isin(0.5tan^{-1}(4/3))$
$z_{2}=sqrt[4]5cdot e^{frac{1}{2}itan^{-1}(4/3)+ipi}\=sqrt[4]5(cos(0.5tan^{-1}(4/3)+pi) + isin(0.5tan^{-1}(4/3)+pi)$
Yet the correct solution is:
$z=pm sqrt[4]5(cos(0.5tan^{-1}(4/3)) + isin(0.5tan^{-1}(4/3))\$
So $z_1$ is a correct solution, but i'm not sure what about $z_2$ and why there is a negative solution in the answer.
Thanks
complex-numbers
complex-numbers
edited Jan 20 at 19:47
J. W. Tanner
2,7911217
asked Jan 20 at 18:31
PulleyBoyPulleyBoy
31
edited Jan 20 at 19:47
J. W. Tanner
2,7911217
asked Jan 20 at 18:31
PulleyBoyPulleyBoy
31
edited Jan 20 at 19:47
J. W. Tanner
2,7911217
edited Jan 20 at 19:47
J. W. Tanner
2,7911217
edited Jan 20 at 19:47
J. W. Tanner
2,7911217
2,7911217
asked Jan 20 at 18:31
PulleyBoyPulleyBoy
31
asked Jan 20 at 18:31
PulleyBoyPulleyBoy
31
asked Jan 20 at 18:31
PulleyBoyPulleyBoy
31
31
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Simplify $z_2$ using $cos(theta+pi)=-cos(theta)$ and $sin(theta+pi)=-sin(theta)$.
answered Jan 20 at 18:35
J. W. TannerJ. W. Tanner
2,7911217
$endgroup$
add a comment |
$begingroup$
You can simplify $z^3bar z=sqrt{5}z^2$ giving $$z^2=dfrac{3+4i}{sqrt{5}}iff z=pmdfrac{2+i}{sqrt[4]{5}}$$
Since $(2+i)^2=4+4i-1=3+4i$
answered Jan 20 at 19:00
zwimzwim
12.5k831
$endgroup$
$begingroup$
Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any?
$endgroup$
– user376343
Jan 20 at 23:36
add a comment |
$begingroup$
From$$r^3rtext{ cis }(3theta-theta)=5text{ cis}arctanfrac43,$$
we draw
$$r=sqrt[4]5,$$
$$theta=frac12arctanfrac43+kpi.$$
Hence
$$pmsqrt[4]5text{ cis}left(frac12arctanfrac43right).$$
(The $pm$ comes from $kpi$.)
answered Jan 20 at 19:16
Yves DaoustYves Daoust
129k676227
$endgroup$
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',
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080956%2fcomplex-equation-with-a-conjugate%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Simplify $z_2$ using $cos(theta+pi)=-cos(theta)$ and $sin(theta+pi)=-sin(theta)$.
answered Jan 20 at 18:35
J. W. TannerJ. W. Tanner
2,7911217
$endgroup$
add a comment |
$begingroup$
Simplify $z_2$ using $cos(theta+pi)=-cos(theta)$ and $sin(theta+pi)=-sin(theta)$.
answered Jan 20 at 18:35
J. W. TannerJ. W. Tanner
2,7911217
$endgroup$
add a comment |
$begingroup$
Simplify $z_2$ using $cos(theta+pi)=-cos(theta)$ and $sin(theta+pi)=-sin(theta)$.
answered Jan 20 at 18:35
J. W. TannerJ. W. Tanner
2,7911217
$endgroup$
Simplify $z_2$ using $cos(theta+pi)=-cos(theta)$ and $sin(theta+pi)=-sin(theta)$.
answered Jan 20 at 18:35
J. W. TannerJ. W. Tanner
2,7911217
edited Jan 20 at 19:05
answered Jan 20 at 18:35
J. W. TannerJ. W. Tanner
2,7911217
answered Jan 20 at 18:35
J. W. TannerJ. W. Tanner
2,7911217
answered Jan 20 at 18:35
J. W. TannerJ. W. Tanner
2,7911217
2,7911217
add a comment |
add a comment |
$begingroup$
You can simplify $z^3bar z=sqrt{5}z^2$ giving $$z^2=dfrac{3+4i}{sqrt{5}}iff z=pmdfrac{2+i}{sqrt[4]{5}}$$
Since $(2+i)^2=4+4i-1=3+4i$
answered Jan 20 at 19:00
zwimzwim
12.5k831
$endgroup$
$begingroup$
Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any?
$endgroup$
– user376343
Jan 20 at 23:36
add a comment |
$begingroup$
You can simplify $z^3bar z=sqrt{5}z^2$ giving $$z^2=dfrac{3+4i}{sqrt{5}}iff z=pmdfrac{2+i}{sqrt[4]{5}}$$
Since $(2+i)^2=4+4i-1=3+4i$
answered Jan 20 at 19:00
zwimzwim
12.5k831
$endgroup$
$begingroup$
Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any?
$endgroup$
– user376343
Jan 20 at 23:36
add a comment |
$begingroup$
You can simplify $z^3bar z=sqrt{5}z^2$ giving $$z^2=dfrac{3+4i}{sqrt{5}}iff z=pmdfrac{2+i}{sqrt[4]{5}}$$
Since $(2+i)^2=4+4i-1=3+4i$
answered Jan 20 at 19:00
zwimzwim
12.5k831
$endgroup$
You can simplify $z^3bar z=sqrt{5}z^2$ giving $$z^2=dfrac{3+4i}{sqrt{5}}iff z=pmdfrac{2+i}{sqrt[4]{5}}$$
Since $(2+i)^2=4+4i-1=3+4i$
answered Jan 20 at 19:00
zwimzwim
12.5k831
answered Jan 20 at 19:00
zwimzwim
12.5k831
answered Jan 20 at 19:00
zwimzwim
12.5k831
answered Jan 20 at 19:00
zwimzwim
12.5k831
12.5k831
$begingroup$
Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any?
$endgroup$
– user376343
Jan 20 at 23:36
add a comment |
$begingroup$
Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any?
$endgroup$
– user376343
Jan 20 at 23:36
$begingroup$
Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any?
$endgroup$
– user376343
Jan 20 at 23:36
$begingroup$
Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any?
$endgroup$
– user376343
Jan 20 at 23:36
add a comment |
$begingroup$
From$$r^3rtext{ cis }(3theta-theta)=5text{ cis}arctanfrac43,$$
we draw
$$r=sqrt[4]5,$$
$$theta=frac12arctanfrac43+kpi.$$
Hence
$$pmsqrt[4]5text{ cis}left(frac12arctanfrac43right).$$
(The $pm$ comes from $kpi$.)
answered Jan 20 at 19:16
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
From$$r^3rtext{ cis }(3theta-theta)=5text{ cis}arctanfrac43,$$
we draw
$$r=sqrt[4]5,$$
$$theta=frac12arctanfrac43+kpi.$$
Hence
$$pmsqrt[4]5text{ cis}left(frac12arctanfrac43right).$$
(The $pm$ comes from $kpi$.)
answered Jan 20 at 19:16
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
From$$r^3rtext{ cis }(3theta-theta)=5text{ cis}arctanfrac43,$$
we draw
$$r=sqrt[4]5,$$
$$theta=frac12arctanfrac43+kpi.$$
Hence
$$pmsqrt[4]5text{ cis}left(frac12arctanfrac43right).$$
(The $pm$ comes from $kpi$.)
answered Jan 20 at 19:16
Yves DaoustYves Daoust
129k676227
$endgroup$
From$$r^3rtext{ cis }(3theta-theta)=5text{ cis}arctanfrac43,$$
we draw
$$r=sqrt[4]5,$$
$$theta=frac12arctanfrac43+kpi.$$
Hence
$$pmsqrt[4]5text{ cis}left(frac12arctanfrac43right).$$
(The $pm$ comes from $kpi$.)
answered Jan 20 at 19:16
Yves DaoustYves Daoust
129k676227
answered Jan 20 at 19:16
Yves DaoustYves Daoust
129k676227
answered Jan 20 at 19:16
Yves DaoustYves Daoust
129k676227
answered Jan 20 at 19:16
Yves DaoustYves Daoust
129k676227
129k676227
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080956%2fcomplex-equation-with-a-conjugate%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
