Mixing
Solving $ z cdot |z| = 4 $
I have to solve $ (z^{*})^2 cdot z^6 = 256 $
After some transformations I have:
$$ |z|cdot z = 4 vee |z|cdot z = -4 vee |z|cdot z = 4i vee |z|cdot z = -4i $$
How can I solve $ |z|cdot z = 4$ ? I have no idea how can I do that...
algebra-precalculus complex-numbers absolute-value
edited Nov 20 '18 at 23:18
Batominovski
33.9k33292
asked Nov 20 '18 at 23:08
VirtualUser
40511
add a comment |
I have to solve $ (z^{*})^2 cdot z^6 = 256 $
After some transformations I have:
$$ |z|cdot z = 4 vee |z|cdot z = -4 vee |z|cdot z = 4i vee |z|cdot z = -4i $$
How can I solve $ |z|cdot z = 4$ ? I have no idea how can I do that...
algebra-precalculus complex-numbers absolute-value
edited Nov 20 '18 at 23:18
Batominovski
33.9k33292
asked Nov 20 '18 at 23:08
VirtualUser
40511
1
Writing this as $z^2,|z|^2=16$ and using $|z|^2=z,bar{z}$, this gives $z^3,bar{z}=16$, so $z^3=16bar{z}^{-1}$. You can use the result from this answer with $m=3$ and $n=-1$. (In this answer $m$ and $n$ are assumed to be positive integers, but nothing changes much if you allow $m$ or $n$ to be nonpositive integers.) Be aware that we have squared the equation, so some of the solutions obtained may satisfy $z,|z|=-4$ instead.
– Batominovski
Nov 20 '18 at 23:14
Oopss, I think I messed up $m$ and $n$. It should be $n=3$ and $m=-1$ in the comment above.
– Batominovski
Nov 20 '18 at 23:20
1
Hmm, I just realized that the title of your question is not the same as the body of your question. If you want to solve $bar{z}^2,z^6=256$, then you can simply use the result from the link I gave in my first answer, using $n=6$ and $m=-2$.
– Batominovski
Nov 20 '18 at 23:23
add a comment |
I have to solve $ (z^{*})^2 cdot z^6 = 256 $
After some transformations I have:
$$ |z|cdot z = 4 vee |z|cdot z = -4 vee |z|cdot z = 4i vee |z|cdot z = -4i $$
How can I solve $ |z|cdot z = 4$ ? I have no idea how can I do that...
algebra-precalculus complex-numbers absolute-value
edited Nov 20 '18 at 23:18
Batominovski
33.9k33292
asked Nov 20 '18 at 23:08
VirtualUser
40511
I have to solve $ (z^{*})^2 cdot z^6 = 256 $
After some transformations I have:
$$ |z|cdot z = 4 vee |z|cdot z = -4 vee |z|cdot z = 4i vee |z|cdot z = -4i $$
How can I solve $ |z|cdot z = 4$ ? I have no idea how can I do that...
algebra-precalculus complex-numbers absolute-value
algebra-precalculus complex-numbers absolute-value
edited Nov 20 '18 at 23:18
Batominovski
33.9k33292
asked Nov 20 '18 at 23:08
VirtualUser
40511
edited Nov 20 '18 at 23:18
Batominovski
33.9k33292
asked Nov 20 '18 at 23:08
VirtualUser
40511
edited Nov 20 '18 at 23:18
Batominovski
33.9k33292
edited Nov 20 '18 at 23:18
Batominovski
33.9k33292
edited Nov 20 '18 at 23:18
Batominovski
33.9k33292
33.9k33292
asked Nov 20 '18 at 23:08
VirtualUser
40511
asked Nov 20 '18 at 23:08
VirtualUser
40511
asked Nov 20 '18 at 23:08
VirtualUser
40511
40511
1
Writing this as $z^2,|z|^2=16$ and using $|z|^2=z,bar{z}$, this gives $z^3,bar{z}=16$, so $z^3=16bar{z}^{-1}$. You can use the result from this answer with $m=3$ and $n=-1$. (In this answer $m$ and $n$ are assumed to be positive integers, but nothing changes much if you allow $m$ or $n$ to be nonpositive integers.) Be aware that we have squared the equation, so some of the solutions obtained may satisfy $z,|z|=-4$ instead.
– Batominovski
Nov 20 '18 at 23:14
Oopss, I think I messed up $m$ and $n$. It should be $n=3$ and $m=-1$ in the comment above.
– Batominovski
Nov 20 '18 at 23:20
1
Hmm, I just realized that the title of your question is not the same as the body of your question. If you want to solve $bar{z}^2,z^6=256$, then you can simply use the result from the link I gave in my first answer, using $n=6$ and $m=-2$.
– Batominovski
Nov 20 '18 at 23:23
add a comment |
1
Writing this as $z^2,|z|^2=16$ and using $|z|^2=z,bar{z}$, this gives $z^3,bar{z}=16$, so $z^3=16bar{z}^{-1}$. You can use the result from this answer with $m=3$ and $n=-1$. (In this answer $m$ and $n$ are assumed to be positive integers, but nothing changes much if you allow $m$ or $n$ to be nonpositive integers.) Be aware that we have squared the equation, so some of the solutions obtained may satisfy $z,|z|=-4$ instead.
– Batominovski
Nov 20 '18 at 23:14
Oopss, I think I messed up $m$ and $n$. It should be $n=3$ and $m=-1$ in the comment above.
– Batominovski
Nov 20 '18 at 23:20
1
Hmm, I just realized that the title of your question is not the same as the body of your question. If you want to solve $bar{z}^2,z^6=256$, then you can simply use the result from the link I gave in my first answer, using $n=6$ and $m=-2$.
– Batominovski
Nov 20 '18 at 23:23
1
1
Writing this as $z^2,|z|^2=16$ and using $|z|^2=z,bar{z}$, this gives $z^3,bar{z}=16$, so $z^3=16bar{z}^{-1}$. You can use the result from this answer with $m=3$ and $n=-1$. (In this answer $m$ and $n$ are assumed to be positive integers, but nothing changes much if you allow $m$ or $n$ to be nonpositive integers.) Be aware that we have squared the equation, so some of the solutions obtained may satisfy $z,|z|=-4$ instead.
– Batominovski
Nov 20 '18 at 23:14
Writing this as $z^2,|z|^2=16$ and using $|z|^2=z,bar{z}$, this gives $z^3,bar{z}=16$, so $z^3=16bar{z}^{-1}$. You can use the result from this answer with $m=3$ and $n=-1$. (In this answer $m$ and $n$ are assumed to be positive integers, but nothing changes much if you allow $m$ or $n$ to be nonpositive integers.) Be aware that we have squared the equation, so some of the solutions obtained may satisfy $z,|z|=-4$ instead.
– Batominovski
Nov 20 '18 at 23:14
Oopss, I think I messed up $m$ and $n$. It should be $n=3$ and $m=-1$ in the comment above.
– Batominovski
Nov 20 '18 at 23:20
Oopss, I think I messed up $m$ and $n$. It should be $n=3$ and $m=-1$ in the comment above.
– Batominovski
Nov 20 '18 at 23:20
1
1
Hmm, I just realized that the title of your question is not the same as the body of your question. If you want to solve $bar{z}^2,z^6=256$, then you can simply use the result from the link I gave in my first answer, using $n=6$ and $m=-2$.
– Batominovski
Nov 20 '18 at 23:23
Hmm, I just realized that the title of your question is not the same as the body of your question. If you want to solve $bar{z}^2,z^6=256$, then you can simply use the result from the link I gave in my first answer, using $n=6$ and $m=-2$.
– Batominovski
Nov 20 '18 at 23:23
add a comment |
3 Answers
3
active
oldest
votes
HINT
We have that
$$(z^{*})^2 cdot z^6 = 256 iff (z^{*})^2 z^2cdot z^4 = 256 iff |z|^4z^4=256 implies |z|=2$$
As an alternative by $z=re^{itheta}$
$$(z^{*})^2 cdot z^6 = 256 iff r^2e^{-i2theta}cdot r^6e^{i6theta}=256$$
answered Nov 20 '18 at 23:12
gimusi
1
In me question I suggested that I don't understand how can I solve $|z|cdot z = 4$ so this is the same problem.
– VirtualUser
Nov 20 '18 at 23:13
@VirtualUser In the same way for $|z|cdot z = 4 implies |z|=2$ then from $|z|^4z^4=256 implies z^4=16$
– gimusi
Nov 20 '18 at 23:16
Ah, I know that I can do this in the same way, but I don't know why is it works?
– VirtualUser
Nov 20 '18 at 23:17
@VirtualUser Let $z=re^{itheta}$ to obtain the same result.
– gimusi
Nov 20 '18 at 23:18
Does it mean that I can put module from both sides?
– VirtualUser
Nov 20 '18 at 23:20
|
show 1 more comment
Addressing strictly the question asked, if $z|z|=4$, then we know that $znot=0$, so that $|z|$ is a positive real number, and hence $z=4/|z|$ is also a positive real number. But that implies $|z|=z$, so the equation $z|z|=4$ becomes $z^2=4$, which gives us $z=sqrt4=2$ (since we've already established that $z$ is positive).
answered Nov 20 '18 at 23:54
Barry Cipra
59.1k653124
add a comment |
$$(z^*)^2times z^6=256$$
$$(z^*)^2times z^2times z^4=256$$
and we know that:
$$(z^*)^2times z^2=|z|^4$$
$$therefore |z|^4times z^4=256$$
$$|z|times z=pm4$$
which is what you got.
now:
$$left|re^{itheta}right|times re^{itheta}=pm4$$
visualising this as coordinates we get:
$$r^2costheta=pm4$$
$$r^2sintheta=0$$
answered Nov 20 '18 at 23:40
Henry Lee
1,773218
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
We have that
$$(z^{*})^2 cdot z^6 = 256 iff (z^{*})^2 z^2cdot z^4 = 256 iff |z|^4z^4=256 implies |z|=2$$
As an alternative by $z=re^{itheta}$
$$(z^{*})^2 cdot z^6 = 256 iff r^2e^{-i2theta}cdot r^6e^{i6theta}=256$$
answered Nov 20 '18 at 23:12
gimusi
1
In me question I suggested that I don't understand how can I solve $|z|cdot z = 4$ so this is the same problem.
– VirtualUser
Nov 20 '18 at 23:13
@VirtualUser In the same way for $|z|cdot z = 4 implies |z|=2$ then from $|z|^4z^4=256 implies z^4=16$
– gimusi
Nov 20 '18 at 23:16
Ah, I know that I can do this in the same way, but I don't know why is it works?
– VirtualUser
Nov 20 '18 at 23:17
@VirtualUser Let $z=re^{itheta}$ to obtain the same result.
– gimusi
Nov 20 '18 at 23:18
Does it mean that I can put module from both sides?
– VirtualUser
Nov 20 '18 at 23:20
|
show 1 more comment
HINT
We have that
$$(z^{*})^2 cdot z^6 = 256 iff (z^{*})^2 z^2cdot z^4 = 256 iff |z|^4z^4=256 implies |z|=2$$
As an alternative by $z=re^{itheta}$
$$(z^{*})^2 cdot z^6 = 256 iff r^2e^{-i2theta}cdot r^6e^{i6theta}=256$$
answered Nov 20 '18 at 23:12
gimusi
1
In me question I suggested that I don't understand how can I solve $|z|cdot z = 4$ so this is the same problem.
– VirtualUser
Nov 20 '18 at 23:13
@VirtualUser In the same way for $|z|cdot z = 4 implies |z|=2$ then from $|z|^4z^4=256 implies z^4=16$
– gimusi
Nov 20 '18 at 23:16
Ah, I know that I can do this in the same way, but I don't know why is it works?
– VirtualUser
Nov 20 '18 at 23:17
@VirtualUser Let $z=re^{itheta}$ to obtain the same result.
– gimusi
Nov 20 '18 at 23:18
Does it mean that I can put module from both sides?
– VirtualUser
Nov 20 '18 at 23:20
|
show 1 more comment
HINT
We have that
$$(z^{*})^2 cdot z^6 = 256 iff (z^{*})^2 z^2cdot z^4 = 256 iff |z|^4z^4=256 implies |z|=2$$
As an alternative by $z=re^{itheta}$
$$(z^{*})^2 cdot z^6 = 256 iff r^2e^{-i2theta}cdot r^6e^{i6theta}=256$$
answered Nov 20 '18 at 23:12
gimusi
1
HINT
We have that
$$(z^{*})^2 cdot z^6 = 256 iff (z^{*})^2 z^2cdot z^4 = 256 iff |z|^4z^4=256 implies |z|=2$$
As an alternative by $z=re^{itheta}$
$$(z^{*})^2 cdot z^6 = 256 iff r^2e^{-i2theta}cdot r^6e^{i6theta}=256$$
answered Nov 20 '18 at 23:12
gimusi
1
edited Nov 20 '18 at 23:20
answered Nov 20 '18 at 23:12
gimusi
1
answered Nov 20 '18 at 23:12
gimusi
1
answered Nov 20 '18 at 23:12
gimusi
1
1
In me question I suggested that I don't understand how can I solve $|z|cdot z = 4$ so this is the same problem.
– VirtualUser
Nov 20 '18 at 23:13
@VirtualUser In the same way for $|z|cdot z = 4 implies |z|=2$ then from $|z|^4z^4=256 implies z^4=16$
– gimusi
Nov 20 '18 at 23:16
Ah, I know that I can do this in the same way, but I don't know why is it works?
– VirtualUser
Nov 20 '18 at 23:17
@VirtualUser Let $z=re^{itheta}$ to obtain the same result.
– gimusi
Nov 20 '18 at 23:18
Does it mean that I can put module from both sides?
– VirtualUser
Nov 20 '18 at 23:20
|
show 1 more comment
In me question I suggested that I don't understand how can I solve $|z|cdot z = 4$ so this is the same problem.
– VirtualUser
Nov 20 '18 at 23:13
@VirtualUser In the same way for $|z|cdot z = 4 implies |z|=2$ then from $|z|^4z^4=256 implies z^4=16$
– gimusi
Nov 20 '18 at 23:16
Ah, I know that I can do this in the same way, but I don't know why is it works?
– VirtualUser
Nov 20 '18 at 23:17
@VirtualUser Let $z=re^{itheta}$ to obtain the same result.
– gimusi
Nov 20 '18 at 23:18
Does it mean that I can put module from both sides?
– VirtualUser
Nov 20 '18 at 23:20
In me question I suggested that I don't understand how can I solve $|z|cdot z = 4$ so this is the same problem.
– VirtualUser
Nov 20 '18 at 23:13
In me question I suggested that I don't understand how can I solve $|z|cdot z = 4$ so this is the same problem.
– VirtualUser
Nov 20 '18 at 23:13
@VirtualUser In the same way for $|z|cdot z = 4 implies |z|=2$ then from $|z|^4z^4=256 implies z^4=16$
– gimusi
Nov 20 '18 at 23:16
@VirtualUser In the same way for $|z|cdot z = 4 implies |z|=2$ then from $|z|^4z^4=256 implies z^4=16$
– gimusi
Nov 20 '18 at 23:16
Ah, I know that I can do this in the same way, but I don't know why is it works?
– VirtualUser
Nov 20 '18 at 23:17
Ah, I know that I can do this in the same way, but I don't know why is it works?
– VirtualUser
Nov 20 '18 at 23:17
@VirtualUser Let $z=re^{itheta}$ to obtain the same result.
– gimusi
Nov 20 '18 at 23:18
@VirtualUser Let $z=re^{itheta}$ to obtain the same result.
– gimusi
Nov 20 '18 at 23:18
Does it mean that I can put module from both sides?
– VirtualUser
Nov 20 '18 at 23:20
Does it mean that I can put module from both sides?
– VirtualUser
Nov 20 '18 at 23:20
|
show 1 more comment
Addressing strictly the question asked, if $z|z|=4$, then we know that $znot=0$, so that $|z|$ is a positive real number, and hence $z=4/|z|$ is also a positive real number. But that implies $|z|=z$, so the equation $z|z|=4$ becomes $z^2=4$, which gives us $z=sqrt4=2$ (since we've already established that $z$ is positive).
answered Nov 20 '18 at 23:54
Barry Cipra
59.1k653124
add a comment |
Addressing strictly the question asked, if $z|z|=4$, then we know that $znot=0$, so that $|z|$ is a positive real number, and hence $z=4/|z|$ is also a positive real number. But that implies $|z|=z$, so the equation $z|z|=4$ becomes $z^2=4$, which gives us $z=sqrt4=2$ (since we've already established that $z$ is positive).
answered Nov 20 '18 at 23:54
Barry Cipra
59.1k653124
add a comment |
Addressing strictly the question asked, if $z|z|=4$, then we know that $znot=0$, so that $|z|$ is a positive real number, and hence $z=4/|z|$ is also a positive real number. But that implies $|z|=z$, so the equation $z|z|=4$ becomes $z^2=4$, which gives us $z=sqrt4=2$ (since we've already established that $z$ is positive).
answered Nov 20 '18 at 23:54
Barry Cipra
59.1k653124
Addressing strictly the question asked, if $z|z|=4$, then we know that $znot=0$, so that $|z|$ is a positive real number, and hence $z=4/|z|$ is also a positive real number. But that implies $|z|=z$, so the equation $z|z|=4$ becomes $z^2=4$, which gives us $z=sqrt4=2$ (since we've already established that $z$ is positive).
answered Nov 20 '18 at 23:54
Barry Cipra
59.1k653124
answered Nov 20 '18 at 23:54
Barry Cipra
59.1k653124
answered Nov 20 '18 at 23:54
Barry Cipra
59.1k653124
answered Nov 20 '18 at 23:54
Barry Cipra
59.1k653124
59.1k653124
add a comment |
add a comment |
$$(z^*)^2times z^6=256$$
$$(z^*)^2times z^2times z^4=256$$
and we know that:
$$(z^*)^2times z^2=|z|^4$$
$$therefore |z|^4times z^4=256$$
$$|z|times z=pm4$$
which is what you got.
now:
$$left|re^{itheta}right|times re^{itheta}=pm4$$
visualising this as coordinates we get:
$$r^2costheta=pm4$$
$$r^2sintheta=0$$
answered Nov 20 '18 at 23:40
Henry Lee
1,773218
add a comment |
$$(z^*)^2times z^6=256$$
$$(z^*)^2times z^2times z^4=256$$
and we know that:
$$(z^*)^2times z^2=|z|^4$$
$$therefore |z|^4times z^4=256$$
$$|z|times z=pm4$$
which is what you got.
now:
$$left|re^{itheta}right|times re^{itheta}=pm4$$
visualising this as coordinates we get:
$$r^2costheta=pm4$$
$$r^2sintheta=0$$
answered Nov 20 '18 at 23:40
Henry Lee
1,773218
add a comment |
$$(z^*)^2times z^6=256$$
$$(z^*)^2times z^2times z^4=256$$
and we know that:
$$(z^*)^2times z^2=|z|^4$$
$$therefore |z|^4times z^4=256$$
$$|z|times z=pm4$$
which is what you got.
now:
$$left|re^{itheta}right|times re^{itheta}=pm4$$
visualising this as coordinates we get:
$$r^2costheta=pm4$$
$$r^2sintheta=0$$
answered Nov 20 '18 at 23:40
Henry Lee
1,773218
$$(z^*)^2times z^6=256$$
$$(z^*)^2times z^2times z^4=256$$
and we know that:
$$(z^*)^2times z^2=|z|^4$$
$$therefore |z|^4times z^4=256$$
$$|z|times z=pm4$$
which is what you got.
now:
$$left|re^{itheta}right|times re^{itheta}=pm4$$
visualising this as coordinates we get:
$$r^2costheta=pm4$$
$$r^2sintheta=0$$
answered Nov 20 '18 at 23:40
Henry Lee
1,773218
answered Nov 20 '18 at 23:40
Henry Lee
1,773218
answered Nov 20 '18 at 23:40
Henry Lee
1,773218
answered Nov 20 '18 at 23:40
Henry Lee
1,773218
1,773218
add a comment |
add a comment |
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1
Writing this as $z^2,|z|^2=16$ and using $|z|^2=z,bar{z}$, this gives $z^3,bar{z}=16$, so $z^3=16bar{z}^{-1}$. You can use the result from this answer with $m=3$ and $n=-1$. (In this answer $m$ and $n$ are assumed to be positive integers, but nothing changes much if you allow $m$ or $n$ to be nonpositive integers.) Be aware that we have squared the equation, so some of the solutions obtained may satisfy $z,|z|=-4$ instead.
– Batominovski
Nov 20 '18 at 23:14
Oopss, I think I messed up $m$ and $n$. It should be $n=3$ and $m=-1$ in the comment above.
– Batominovski
Nov 20 '18 at 23:20
1
Hmm, I just realized that the title of your question is not the same as the body of your question. If you want to solve $bar{z}^2,z^6=256$, then you can simply use the result from the link I gave in my first answer, using $n=6$ and $m=-2$.
– Batominovski
Nov 20 '18 at 23:23