Mixing
Finding all roots of equations which solutions are different than $z=0$
$$2overline z = z^7 , , ,,32overline z = z^7 ,,,, 128overline z + z^7=0 $$
Let's say that $z=r\$ then $2r=r^7$ then $2=r^6$ which gives us $r=sqrt[6]{2}$.
Now we take (argument) $$text{arg}, z=alpha,, ,,text{arg},overline z=-alpha, , ,,text{arg},2overline z=-alpha$$ so $$-alpha=7 alpha , , ,,8alpha=0,,,,8alpha=2kpi,,,, alpha=kfrac {pi}{4}.$$
Am I doing this the correct way? Can someone explain what am I supposed to do next? How can I solve other equations?
algebra-precalculus proof-verification complex-numbers
edited Nov 20 '18 at 20:38
Batominovski
33.8k33292
asked Nov 20 '18 at 17:18
B. Czostek
294
add a comment |
$$2overline z = z^7 , , ,,32overline z = z^7 ,,,, 128overline z + z^7=0 $$
Let's say that $z=r\$ then $2r=r^7$ then $2=r^6$ which gives us $r=sqrt[6]{2}$.
Now we take (argument) $$text{arg}, z=alpha,, ,,text{arg},overline z=-alpha, , ,,text{arg},2overline z=-alpha$$ so $$-alpha=7 alpha , , ,,8alpha=0,,,,8alpha=2kpi,,,, alpha=kfrac {pi}{4}.$$
Am I doing this the correct way? Can someone explain what am I supposed to do next? How can I solve other equations?
algebra-precalculus proof-verification complex-numbers
edited Nov 20 '18 at 20:38
Batominovski
33.8k33292
asked Nov 20 '18 at 17:18
B. Czostek
294
Your question is unclear. Do you want to solve the system of equations $2bar z = z^7, 32bar z = z^7$ and $128bar z + z^7=0$ or do you want to solve the equations separately?
– user1551
Nov 20 '18 at 17:26
Solve the equations sparately, but I also want to know how (if they are) are they related to each other when it comes to finding the solution
– B. Czostek
Nov 20 '18 at 17:28
add a comment |
$$2overline z = z^7 , , ,,32overline z = z^7 ,,,, 128overline z + z^7=0 $$
Let's say that $z=r\$ then $2r=r^7$ then $2=r^6$ which gives us $r=sqrt[6]{2}$.
Now we take (argument) $$text{arg}, z=alpha,, ,,text{arg},overline z=-alpha, , ,,text{arg},2overline z=-alpha$$ so $$-alpha=7 alpha , , ,,8alpha=0,,,,8alpha=2kpi,,,, alpha=kfrac {pi}{4}.$$
Am I doing this the correct way? Can someone explain what am I supposed to do next? How can I solve other equations?
algebra-precalculus proof-verification complex-numbers
edited Nov 20 '18 at 20:38
Batominovski
33.8k33292
asked Nov 20 '18 at 17:18
B. Czostek
294
$$2overline z = z^7 , , ,,32overline z = z^7 ,,,, 128overline z + z^7=0 $$
Let's say that $z=r\$ then $2r=r^7$ then $2=r^6$ which gives us $r=sqrt[6]{2}$.
Now we take (argument) $$text{arg}, z=alpha,, ,,text{arg},overline z=-alpha, , ,,text{arg},2overline z=-alpha$$ so $$-alpha=7 alpha , , ,,8alpha=0,,,,8alpha=2kpi,,,, alpha=kfrac {pi}{4}.$$
Am I doing this the correct way? Can someone explain what am I supposed to do next? How can I solve other equations?
algebra-precalculus proof-verification complex-numbers
algebra-precalculus proof-verification complex-numbers
edited Nov 20 '18 at 20:38
Batominovski
33.8k33292
asked Nov 20 '18 at 17:18
B. Czostek
294
edited Nov 20 '18 at 20:38
Batominovski
33.8k33292
asked Nov 20 '18 at 17:18
B. Czostek
294
edited Nov 20 '18 at 20:38
Batominovski
33.8k33292
edited Nov 20 '18 at 20:38
Batominovski
33.8k33292
edited Nov 20 '18 at 20:38
Batominovski
33.8k33292
33.8k33292
asked Nov 20 '18 at 17:18
B. Czostek
294
asked Nov 20 '18 at 17:18
B. Czostek
294
asked Nov 20 '18 at 17:18
B. Czostek
294
294
Your question is unclear. Do you want to solve the system of equations $2bar z = z^7, 32bar z = z^7$ and $128bar z + z^7=0$ or do you want to solve the equations separately?
– user1551
Nov 20 '18 at 17:26
Solve the equations sparately, but I also want to know how (if they are) are they related to each other when it comes to finding the solution
– B. Czostek
Nov 20 '18 at 17:28
add a comment |
Your question is unclear. Do you want to solve the system of equations $2bar z = z^7, 32bar z = z^7$ and $128bar z + z^7=0$ or do you want to solve the equations separately?
– user1551
Nov 20 '18 at 17:26
Solve the equations sparately, but I also want to know how (if they are) are they related to each other when it comes to finding the solution
– B. Czostek
Nov 20 '18 at 17:28
Your question is unclear. Do you want to solve the system of equations $2bar z = z^7, 32bar z = z^7$ and $128bar z + z^7=0$ or do you want to solve the equations separately?
– user1551
Nov 20 '18 at 17:26
Your question is unclear. Do you want to solve the system of equations $2bar z = z^7, 32bar z = z^7$ and $128bar z + z^7=0$ or do you want to solve the equations separately?
– user1551
Nov 20 '18 at 17:26
Solve the equations sparately, but I also want to know how (if they are) are they related to each other when it comes to finding the solution
– B. Czostek
Nov 20 '18 at 17:28
Solve the equations sparately, but I also want to know how (if they are) are they related to each other when it comes to finding the solution
– B. Czostek
Nov 20 '18 at 17:28
add a comment |
3 Answers
3
active
oldest
votes
Let me try to solve $$z^n = tbar{z}^m$$ ($m$ and $n$ are positive integers). If $t=0$, the only solution is $z=0$. Suppose that $tne 0$. Clearly, $z=0$ is a still solution. We are seeking all solutions $zneq 0$.
If $m=n$, we have
$$left(frac{z}{bar{z}}right)^n=t.$$
Taking modulus we get $|t|=1$ is the only way to have a non-zero soln. If $t=e^{theta i}$ for some $thetain [0,2pi)$. That is, $$frac{z}{bar{z}}=e^{theta i+frac{2pi k i}{n}}$$
for some $k=0,1,dots,n-1$. That is, $$z=r e^{frac{theta i}{2}+frac{pi k i}{n}}$$ for some $k=0,1,dots,n-1$ and $rinBbb{R}$ s.t. $rne 0$.
If $mneq n$, we have $$|z|^{n-m}=left|frac{z^n}{bar{z}^m}right|=|t|.$$
So, $|z|=|t|^{frac{1}{n-m}}$. That is,
$$z=|t|^{frac{1}{n-m}}e^{phi i}$$
for some $phiin[0,2pi)$. Plugging this into $z^n=tbar{z}^m$, we have
$$|t|^{frac{n}{n-m}}e^{nphi i}=t|t|^{frac{m}{n-m}}e^{-mphi i}.$$
Let $t=|t|e^{theta i}$ for some $thetain[0,2pi)$. We get
$$e^{(m+n)phi i}=e^{theta i}.$$
So, $phi = frac{theta}{m+n}+frac{2kpi}{m+n}$ for $k=0,1,dots,m+n-1$. I then get
$$z=|t|^{frac{1}{n-m}}e^{frac{theta i}{m+n}+frac{2kpi i}{m+n}}$$
for $k=0,1,dots,m+n-1$.
answered Nov 20 '18 at 17:38
Snookie
1,30017
2
If $m$ and $n$ can be nonpositive intgers, then the work is not much different. The only exceptional case is when $m+n=0$, which offers a solution only when $tinmathbb{R}_{>0}$.
– Batominovski
Nov 20 '18 at 20:45
add a comment |
HINT
Multiplying by $z$ we obtain
$$ 2overline zz = z^8 iff z^8=2|z|^2 implies |z|^6=2 implies |z|=2^frac16$$
and then
$$z^8=2^frac43 implies z=ldots$$
answered Nov 20 '18 at 17:23
gimusi
1
Where did you get 2^4/3 from?
– B. Czostek
Nov 20 '18 at 17:25
We have $|z|=2^frac16$ and from $z^8=2|z|^2 $ we obtain the equation.
– gimusi
Nov 20 '18 at 17:26
So z is equal to the eight root of 2^4/3? I really don't get it
– B. Czostek
Nov 20 '18 at 17:38
Assume $z=re^{itheta}$ then the equation $2overline z = z^7$ corresponds to $$2re^{-itheta} =r^7e^{i7theta} iff r^6e^{i8theta}=2$$ From here we find that $r^6=2$ and $e^{i8theta}=1$ that is $$(z/r)^8=1 iff z^8=r^8=2^frac43$$ That's another equivalent way to get the same equation.
– gimusi
Nov 20 '18 at 17:44
add a comment |
All equations are of the form
$$aoverline z=z^7$$
with $a$ real.
Switching to polar coordinates,
$$|a|r=r^7$$ and $$theta_a-theta_r=7theta_r+2kpi$$ where $theta_a$ is $0$ or $pi$.
Then
$$z=sqrt[6]{|a|}text{ cis}frac{theta_a+2kpi}8.$$
answered Nov 20 '18 at 20:48
Yves Daoust
124k671221
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%2f3006609%2ffinding-all-roots-of-equations-which-solutions-are-different-than-z-0%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
Let me try to solve $$z^n = tbar{z}^m$$ ($m$ and $n$ are positive integers). If $t=0$, the only solution is $z=0$. Suppose that $tne 0$. Clearly, $z=0$ is a still solution. We are seeking all solutions $zneq 0$.
If $m=n$, we have
$$left(frac{z}{bar{z}}right)^n=t.$$
Taking modulus we get $|t|=1$ is the only way to have a non-zero soln. If $t=e^{theta i}$ for some $thetain [0,2pi)$. That is, $$frac{z}{bar{z}}=e^{theta i+frac{2pi k i}{n}}$$
for some $k=0,1,dots,n-1$. That is, $$z=r e^{frac{theta i}{2}+frac{pi k i}{n}}$$ for some $k=0,1,dots,n-1$ and $rinBbb{R}$ s.t. $rne 0$.
If $mneq n$, we have $$|z|^{n-m}=left|frac{z^n}{bar{z}^m}right|=|t|.$$
So, $|z|=|t|^{frac{1}{n-m}}$. That is,
$$z=|t|^{frac{1}{n-m}}e^{phi i}$$
for some $phiin[0,2pi)$. Plugging this into $z^n=tbar{z}^m$, we have
$$|t|^{frac{n}{n-m}}e^{nphi i}=t|t|^{frac{m}{n-m}}e^{-mphi i}.$$
Let $t=|t|e^{theta i}$ for some $thetain[0,2pi)$. We get
$$e^{(m+n)phi i}=e^{theta i}.$$
So, $phi = frac{theta}{m+n}+frac{2kpi}{m+n}$ for $k=0,1,dots,m+n-1$. I then get
$$z=|t|^{frac{1}{n-m}}e^{frac{theta i}{m+n}+frac{2kpi i}{m+n}}$$
for $k=0,1,dots,m+n-1$.
answered Nov 20 '18 at 17:38
Snookie
1,30017
2
If $m$ and $n$ can be nonpositive intgers, then the work is not much different. The only exceptional case is when $m+n=0$, which offers a solution only when $tinmathbb{R}_{>0}$.
– Batominovski
Nov 20 '18 at 20:45
add a comment |
Let me try to solve $$z^n = tbar{z}^m$$ ($m$ and $n$ are positive integers). If $t=0$, the only solution is $z=0$. Suppose that $tne 0$. Clearly, $z=0$ is a still solution. We are seeking all solutions $zneq 0$.
If $m=n$, we have
$$left(frac{z}{bar{z}}right)^n=t.$$
Taking modulus we get $|t|=1$ is the only way to have a non-zero soln. If $t=e^{theta i}$ for some $thetain [0,2pi)$. That is, $$frac{z}{bar{z}}=e^{theta i+frac{2pi k i}{n}}$$
for some $k=0,1,dots,n-1$. That is, $$z=r e^{frac{theta i}{2}+frac{pi k i}{n}}$$ for some $k=0,1,dots,n-1$ and $rinBbb{R}$ s.t. $rne 0$.
If $mneq n$, we have $$|z|^{n-m}=left|frac{z^n}{bar{z}^m}right|=|t|.$$
So, $|z|=|t|^{frac{1}{n-m}}$. That is,
$$z=|t|^{frac{1}{n-m}}e^{phi i}$$
for some $phiin[0,2pi)$. Plugging this into $z^n=tbar{z}^m$, we have
$$|t|^{frac{n}{n-m}}e^{nphi i}=t|t|^{frac{m}{n-m}}e^{-mphi i}.$$
Let $t=|t|e^{theta i}$ for some $thetain[0,2pi)$. We get
$$e^{(m+n)phi i}=e^{theta i}.$$
So, $phi = frac{theta}{m+n}+frac{2kpi}{m+n}$ for $k=0,1,dots,m+n-1$. I then get
$$z=|t|^{frac{1}{n-m}}e^{frac{theta i}{m+n}+frac{2kpi i}{m+n}}$$
for $k=0,1,dots,m+n-1$.
answered Nov 20 '18 at 17:38
Snookie
1,30017
2
If $m$ and $n$ can be nonpositive intgers, then the work is not much different. The only exceptional case is when $m+n=0$, which offers a solution only when $tinmathbb{R}_{>0}$.
– Batominovski
Nov 20 '18 at 20:45
add a comment |
Let me try to solve $$z^n = tbar{z}^m$$ ($m$ and $n$ are positive integers). If $t=0$, the only solution is $z=0$. Suppose that $tne 0$. Clearly, $z=0$ is a still solution. We are seeking all solutions $zneq 0$.
If $m=n$, we have
$$left(frac{z}{bar{z}}right)^n=t.$$
Taking modulus we get $|t|=1$ is the only way to have a non-zero soln. If $t=e^{theta i}$ for some $thetain [0,2pi)$. That is, $$frac{z}{bar{z}}=e^{theta i+frac{2pi k i}{n}}$$
for some $k=0,1,dots,n-1$. That is, $$z=r e^{frac{theta i}{2}+frac{pi k i}{n}}$$ for some $k=0,1,dots,n-1$ and $rinBbb{R}$ s.t. $rne 0$.
If $mneq n$, we have $$|z|^{n-m}=left|frac{z^n}{bar{z}^m}right|=|t|.$$
So, $|z|=|t|^{frac{1}{n-m}}$. That is,
$$z=|t|^{frac{1}{n-m}}e^{phi i}$$
for some $phiin[0,2pi)$. Plugging this into $z^n=tbar{z}^m$, we have
$$|t|^{frac{n}{n-m}}e^{nphi i}=t|t|^{frac{m}{n-m}}e^{-mphi i}.$$
Let $t=|t|e^{theta i}$ for some $thetain[0,2pi)$. We get
$$e^{(m+n)phi i}=e^{theta i}.$$
So, $phi = frac{theta}{m+n}+frac{2kpi}{m+n}$ for $k=0,1,dots,m+n-1$. I then get
$$z=|t|^{frac{1}{n-m}}e^{frac{theta i}{m+n}+frac{2kpi i}{m+n}}$$
for $k=0,1,dots,m+n-1$.
answered Nov 20 '18 at 17:38
Snookie
1,30017
Let me try to solve $$z^n = tbar{z}^m$$ ($m$ and $n$ are positive integers). If $t=0$, the only solution is $z=0$. Suppose that $tne 0$. Clearly, $z=0$ is a still solution. We are seeking all solutions $zneq 0$.
If $m=n$, we have
$$left(frac{z}{bar{z}}right)^n=t.$$
Taking modulus we get $|t|=1$ is the only way to have a non-zero soln. If $t=e^{theta i}$ for some $thetain [0,2pi)$. That is, $$frac{z}{bar{z}}=e^{theta i+frac{2pi k i}{n}}$$
for some $k=0,1,dots,n-1$. That is, $$z=r e^{frac{theta i}{2}+frac{pi k i}{n}}$$ for some $k=0,1,dots,n-1$ and $rinBbb{R}$ s.t. $rne 0$.
If $mneq n$, we have $$|z|^{n-m}=left|frac{z^n}{bar{z}^m}right|=|t|.$$
So, $|z|=|t|^{frac{1}{n-m}}$. That is,
$$z=|t|^{frac{1}{n-m}}e^{phi i}$$
for some $phiin[0,2pi)$. Plugging this into $z^n=tbar{z}^m$, we have
$$|t|^{frac{n}{n-m}}e^{nphi i}=t|t|^{frac{m}{n-m}}e^{-mphi i}.$$
Let $t=|t|e^{theta i}$ for some $thetain[0,2pi)$. We get
$$e^{(m+n)phi i}=e^{theta i}.$$
So, $phi = frac{theta}{m+n}+frac{2kpi}{m+n}$ for $k=0,1,dots,m+n-1$. I then get
$$z=|t|^{frac{1}{n-m}}e^{frac{theta i}{m+n}+frac{2kpi i}{m+n}}$$
for $k=0,1,dots,m+n-1$.
answered Nov 20 '18 at 17:38
Snookie
1,30017
answered Nov 20 '18 at 17:38
Snookie
1,30017
answered Nov 20 '18 at 17:38
Snookie
1,30017
answered Nov 20 '18 at 17:38
Snookie
1,30017
1,30017
2
If $m$ and $n$ can be nonpositive intgers, then the work is not much different. The only exceptional case is when $m+n=0$, which offers a solution only when $tinmathbb{R}_{>0}$.
– Batominovski
Nov 20 '18 at 20:45
add a comment |
2
If $m$ and $n$ can be nonpositive intgers, then the work is not much different. The only exceptional case is when $m+n=0$, which offers a solution only when $tinmathbb{R}_{>0}$.
– Batominovski
Nov 20 '18 at 20:45
2
2
If $m$ and $n$ can be nonpositive intgers, then the work is not much different. The only exceptional case is when $m+n=0$, which offers a solution only when $tinmathbb{R}_{>0}$.
– Batominovski
Nov 20 '18 at 20:45
If $m$ and $n$ can be nonpositive intgers, then the work is not much different. The only exceptional case is when $m+n=0$, which offers a solution only when $tinmathbb{R}_{>0}$.
– Batominovski
Nov 20 '18 at 20:45
add a comment |
HINT
Multiplying by $z$ we obtain
$$ 2overline zz = z^8 iff z^8=2|z|^2 implies |z|^6=2 implies |z|=2^frac16$$
and then
$$z^8=2^frac43 implies z=ldots$$
answered Nov 20 '18 at 17:23
gimusi
1
Where did you get 2^4/3 from?
– B. Czostek
Nov 20 '18 at 17:25
We have $|z|=2^frac16$ and from $z^8=2|z|^2 $ we obtain the equation.
– gimusi
Nov 20 '18 at 17:26
So z is equal to the eight root of 2^4/3? I really don't get it
– B. Czostek
Nov 20 '18 at 17:38
Assume $z=re^{itheta}$ then the equation $2overline z = z^7$ corresponds to $$2re^{-itheta} =r^7e^{i7theta} iff r^6e^{i8theta}=2$$ From here we find that $r^6=2$ and $e^{i8theta}=1$ that is $$(z/r)^8=1 iff z^8=r^8=2^frac43$$ That's another equivalent way to get the same equation.
– gimusi
Nov 20 '18 at 17:44
add a comment |
HINT
Multiplying by $z$ we obtain
$$ 2overline zz = z^8 iff z^8=2|z|^2 implies |z|^6=2 implies |z|=2^frac16$$
and then
$$z^8=2^frac43 implies z=ldots$$
answered Nov 20 '18 at 17:23
gimusi
1
Where did you get 2^4/3 from?
– B. Czostek
Nov 20 '18 at 17:25
We have $|z|=2^frac16$ and from $z^8=2|z|^2 $ we obtain the equation.
– gimusi
Nov 20 '18 at 17:26
So z is equal to the eight root of 2^4/3? I really don't get it
– B. Czostek
Nov 20 '18 at 17:38
Assume $z=re^{itheta}$ then the equation $2overline z = z^7$ corresponds to $$2re^{-itheta} =r^7e^{i7theta} iff r^6e^{i8theta}=2$$ From here we find that $r^6=2$ and $e^{i8theta}=1$ that is $$(z/r)^8=1 iff z^8=r^8=2^frac43$$ That's another equivalent way to get the same equation.
– gimusi
Nov 20 '18 at 17:44
add a comment |
HINT
Multiplying by $z$ we obtain
$$ 2overline zz = z^8 iff z^8=2|z|^2 implies |z|^6=2 implies |z|=2^frac16$$
and then
$$z^8=2^frac43 implies z=ldots$$
answered Nov 20 '18 at 17:23
gimusi
1
HINT
Multiplying by $z$ we obtain
$$ 2overline zz = z^8 iff z^8=2|z|^2 implies |z|^6=2 implies |z|=2^frac16$$
and then
$$z^8=2^frac43 implies z=ldots$$
answered Nov 20 '18 at 17:23
gimusi
1
answered Nov 20 '18 at 17:23
gimusi
1
answered Nov 20 '18 at 17:23
gimusi
1
answered Nov 20 '18 at 17:23
gimusi
1
1
Where did you get 2^4/3 from?
– B. Czostek
Nov 20 '18 at 17:25
We have $|z|=2^frac16$ and from $z^8=2|z|^2 $ we obtain the equation.
– gimusi
Nov 20 '18 at 17:26
So z is equal to the eight root of 2^4/3? I really don't get it
– B. Czostek
Nov 20 '18 at 17:38
Assume $z=re^{itheta}$ then the equation $2overline z = z^7$ corresponds to $$2re^{-itheta} =r^7e^{i7theta} iff r^6e^{i8theta}=2$$ From here we find that $r^6=2$ and $e^{i8theta}=1$ that is $$(z/r)^8=1 iff z^8=r^8=2^frac43$$ That's another equivalent way to get the same equation.
– gimusi
Nov 20 '18 at 17:44
add a comment |
Where did you get 2^4/3 from?
– B. Czostek
Nov 20 '18 at 17:25
We have $|z|=2^frac16$ and from $z^8=2|z|^2 $ we obtain the equation.
– gimusi
Nov 20 '18 at 17:26
So z is equal to the eight root of 2^4/3? I really don't get it
– B. Czostek
Nov 20 '18 at 17:38
Assume $z=re^{itheta}$ then the equation $2overline z = z^7$ corresponds to $$2re^{-itheta} =r^7e^{i7theta} iff r^6e^{i8theta}=2$$ From here we find that $r^6=2$ and $e^{i8theta}=1$ that is $$(z/r)^8=1 iff z^8=r^8=2^frac43$$ That's another equivalent way to get the same equation.
– gimusi
Nov 20 '18 at 17:44
Where did you get 2^4/3 from?
– B. Czostek
Nov 20 '18 at 17:25
Where did you get 2^4/3 from?
– B. Czostek
Nov 20 '18 at 17:25
We have $|z|=2^frac16$ and from $z^8=2|z|^2 $ we obtain the equation.
– gimusi
Nov 20 '18 at 17:26
We have $|z|=2^frac16$ and from $z^8=2|z|^2 $ we obtain the equation.
– gimusi
Nov 20 '18 at 17:26
So z is equal to the eight root of 2^4/3? I really don't get it
– B. Czostek
Nov 20 '18 at 17:38
So z is equal to the eight root of 2^4/3? I really don't get it
– B. Czostek
Nov 20 '18 at 17:38
Assume $z=re^{itheta}$ then the equation $2overline z = z^7$ corresponds to $$2re^{-itheta} =r^7e^{i7theta} iff r^6e^{i8theta}=2$$ From here we find that $r^6=2$ and $e^{i8theta}=1$ that is $$(z/r)^8=1 iff z^8=r^8=2^frac43$$ That's another equivalent way to get the same equation.
– gimusi
Nov 20 '18 at 17:44
Assume $z=re^{itheta}$ then the equation $2overline z = z^7$ corresponds to $$2re^{-itheta} =r^7e^{i7theta} iff r^6e^{i8theta}=2$$ From here we find that $r^6=2$ and $e^{i8theta}=1$ that is $$(z/r)^8=1 iff z^8=r^8=2^frac43$$ That's another equivalent way to get the same equation.
– gimusi
Nov 20 '18 at 17:44
add a comment |
All equations are of the form
$$aoverline z=z^7$$
with $a$ real.
Switching to polar coordinates,
$$|a|r=r^7$$ and $$theta_a-theta_r=7theta_r+2kpi$$ where $theta_a$ is $0$ or $pi$.
Then
$$z=sqrt[6]{|a|}text{ cis}frac{theta_a+2kpi}8.$$
answered Nov 20 '18 at 20:48
Yves Daoust
124k671221
add a comment |
All equations are of the form
$$aoverline z=z^7$$
with $a$ real.
Switching to polar coordinates,
$$|a|r=r^7$$ and $$theta_a-theta_r=7theta_r+2kpi$$ where $theta_a$ is $0$ or $pi$.
Then
$$z=sqrt[6]{|a|}text{ cis}frac{theta_a+2kpi}8.$$
answered Nov 20 '18 at 20:48
Yves Daoust
124k671221
add a comment |
All equations are of the form
$$aoverline z=z^7$$
with $a$ real.
Switching to polar coordinates,
$$|a|r=r^7$$ and $$theta_a-theta_r=7theta_r+2kpi$$ where $theta_a$ is $0$ or $pi$.
Then
$$z=sqrt[6]{|a|}text{ cis}frac{theta_a+2kpi}8.$$
answered Nov 20 '18 at 20:48
Yves Daoust
124k671221
All equations are of the form
$$aoverline z=z^7$$
with $a$ real.
Switching to polar coordinates,
$$|a|r=r^7$$ and $$theta_a-theta_r=7theta_r+2kpi$$ where $theta_a$ is $0$ or $pi$.
Then
$$z=sqrt[6]{|a|}text{ cis}frac{theta_a+2kpi}8.$$
answered Nov 20 '18 at 20:48
Yves Daoust
124k671221
answered Nov 20 '18 at 20:48
Yves Daoust
124k671221
answered Nov 20 '18 at 20:48
Yves Daoust
124k671221
answered Nov 20 '18 at 20:48
Yves Daoust
124k671221
124k671221
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3006609%2ffinding-all-roots-of-equations-which-solutions-are-different-than-z-0%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
Your question is unclear. Do you want to solve the system of equations $2bar z = z^7, 32bar z = z^7$ and $128bar z + z^7=0$ or do you want to solve the equations separately?
– user1551
Nov 20 '18 at 17:26
Solve the equations sparately, but I also want to know how (if they are) are they related to each other when it comes to finding the solution
– B. Czostek
Nov 20 '18 at 17:28