How should I determine all numbers $zin mathbb C$ such that $z^3 = 4overline z$?
$begingroup$
Find all $zin mathbb C$ such that $z^3 = 4overline z$
I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?
complex-numbers
$endgroup$
add a comment |
$begingroup$
Find all $zin mathbb C$ such that $z^3 = 4overline z$
I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?
complex-numbers
$endgroup$
1
$begingroup$
Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
$endgroup$
– lulu
Dec 3 '18 at 0:31
add a comment |
$begingroup$
Find all $zin mathbb C$ such that $z^3 = 4overline z$
I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?
complex-numbers
$endgroup$
Find all $zin mathbb C$ such that $z^3 = 4overline z$
I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?
complex-numbers
complex-numbers
edited Dec 3 '18 at 0:33
lulu
39.7k24677
39.7k24677
asked Dec 3 '18 at 0:23
ReyRey
12
12
1
$begingroup$
Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
$endgroup$
– lulu
Dec 3 '18 at 0:31
add a comment |
1
$begingroup$
Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
$endgroup$
– lulu
Dec 3 '18 at 0:31
1
1
$begingroup$
Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
$endgroup$
– lulu
Dec 3 '18 at 0:31
$begingroup$
Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
$endgroup$
– lulu
Dec 3 '18 at 0:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.
Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.
$endgroup$
$begingroup$
"There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
$endgroup$
– fleablood
Dec 3 '18 at 1:08
add a comment |
$begingroup$
choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.
You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$
or
$r^3 e^{i3theta} = 4re^{-itheta}$
So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$
$r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.
If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$
And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.
And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.
And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.
===
For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so
$x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$
$x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$
$x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$
Case 1:$x = 0$ then $-y^3 = y$
Case 1a: $x = 0$ and $y=0$ and $z = 0$.
Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.
So $z = 2i$ or $z = -2i$
Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$
Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.
So $z = 2$ or $z = -2$.
Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.
So solutions are ${0,2,2i,-2,-2i}$
that actually wasn't so bad. But usually it'd be a lot harder.
=====
I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.
And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.
$endgroup$
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%2f3023424%2fhow-should-i-determine-all-numbers-z-in-mathbb-c-such-that-z3-4-overline%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.
Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.
$endgroup$
$begingroup$
"There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
$endgroup$
– fleablood
Dec 3 '18 at 1:08
add a comment |
$begingroup$
There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.
Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.
$endgroup$
$begingroup$
"There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
$endgroup$
– fleablood
Dec 3 '18 at 1:08
add a comment |
$begingroup$
There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.
Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.
$endgroup$
There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.
Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.
answered Dec 3 '18 at 0:49
Barry CipraBarry Cipra
59.4k653125
59.4k653125
$begingroup$
"There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
$endgroup$
– fleablood
Dec 3 '18 at 1:08
add a comment |
$begingroup$
"There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
$endgroup$
– fleablood
Dec 3 '18 at 1:08
$begingroup$
"There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
$endgroup$
– fleablood
Dec 3 '18 at 1:08
$begingroup$
"There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
$endgroup$
– fleablood
Dec 3 '18 at 1:08
add a comment |
$begingroup$
choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.
You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$
or
$r^3 e^{i3theta} = 4re^{-itheta}$
So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$
$r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.
If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$
And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.
And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.
And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.
===
For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so
$x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$
$x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$
$x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$
Case 1:$x = 0$ then $-y^3 = y$
Case 1a: $x = 0$ and $y=0$ and $z = 0$.
Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.
So $z = 2i$ or $z = -2i$
Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$
Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.
So $z = 2$ or $z = -2$.
Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.
So solutions are ${0,2,2i,-2,-2i}$
that actually wasn't so bad. But usually it'd be a lot harder.
=====
I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.
And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.
$endgroup$
add a comment |
$begingroup$
choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.
You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$
or
$r^3 e^{i3theta} = 4re^{-itheta}$
So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$
$r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.
If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$
And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.
And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.
And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.
===
For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so
$x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$
$x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$
$x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$
Case 1:$x = 0$ then $-y^3 = y$
Case 1a: $x = 0$ and $y=0$ and $z = 0$.
Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.
So $z = 2i$ or $z = -2i$
Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$
Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.
So $z = 2$ or $z = -2$.
Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.
So solutions are ${0,2,2i,-2,-2i}$
that actually wasn't so bad. But usually it'd be a lot harder.
=====
I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.
And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.
$endgroup$
add a comment |
$begingroup$
choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.
You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$
or
$r^3 e^{i3theta} = 4re^{-itheta}$
So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$
$r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.
If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$
And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.
And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.
And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.
===
For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so
$x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$
$x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$
$x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$
Case 1:$x = 0$ then $-y^3 = y$
Case 1a: $x = 0$ and $y=0$ and $z = 0$.
Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.
So $z = 2i$ or $z = -2i$
Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$
Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.
So $z = 2$ or $z = -2$.
Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.
So solutions are ${0,2,2i,-2,-2i}$
that actually wasn't so bad. But usually it'd be a lot harder.
=====
I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.
And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.
$endgroup$
choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.
You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$
or
$r^3 e^{i3theta} = 4re^{-itheta}$
So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$
$r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.
If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$
And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.
And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.
And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.
===
For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so
$x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$
$x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$
$x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$
Case 1:$x = 0$ then $-y^3 = y$
Case 1a: $x = 0$ and $y=0$ and $z = 0$.
Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.
So $z = 2i$ or $z = -2i$
Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$
Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.
So $z = 2$ or $z = -2$.
Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.
So solutions are ${0,2,2i,-2,-2i}$
that actually wasn't so bad. But usually it'd be a lot harder.
=====
I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.
And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.
edited Dec 3 '18 at 1:12
answered Dec 3 '18 at 0:38
fleabloodfleablood
68.9k22685
68.9k22685
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%2f3023424%2fhow-should-i-determine-all-numbers-z-in-mathbb-c-such-that-z3-4-overline%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
1
$begingroup$
Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
$endgroup$
– lulu
Dec 3 '18 at 0:31