How to solve $(1+z^2)^3 = -8$? [on hold]
up vote
0
down vote
favorite
I'm having a hard time understanding how to determine all the complex roots to $(1+z^2)^3 = -8$.
If anyone could help me, I would be so grateful!
complex-numbers
edited yesterday
Florian
1,3001719
asked yesterday
Fosorf
112
New contributor
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, amWhy, Micah, Chris Custer, user10354138 14 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, amWhy, Micah, Chris Custer, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
I'm having a hard time understanding how to determine all the complex roots to $(1+z^2)^3 = -8$.
If anyone could help me, I would be so grateful!
complex-numbers
edited yesterday
Florian
1,3001719
asked yesterday
Fosorf
112
New contributor
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, amWhy, Micah, Chris Custer, user10354138 14 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, amWhy, Micah, Chris Custer, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
3
Would you be able to solve $x^3=-8$ instead? If so, try substituting $1+z^2 = x$. If not, this should be your question.
– Florian
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm having a hard time understanding how to determine all the complex roots to $(1+z^2)^3 = -8$.
If anyone could help me, I would be so grateful!
complex-numbers
edited yesterday
Florian
1,3001719
asked yesterday
Fosorf
112
New contributor
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm having a hard time understanding how to determine all the complex roots to $(1+z^2)^3 = -8$.
If anyone could help me, I would be so grateful!
complex-numbers
complex-numbers
edited yesterday
Florian
1,3001719
asked yesterday
Fosorf
112
New contributor
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Florian
1,3001719
asked yesterday
Fosorf
112
New contributor
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Florian
1,3001719
edited yesterday
Florian
1,3001719
edited yesterday
Florian
1,3001719
1,3001719
asked yesterday
Fosorf
112
New contributor
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Fosorf
112
asked yesterday
Fosorf
112
112
New contributor
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Fosorf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, amWhy, Micah, Chris Custer, user10354138 14 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, amWhy, Micah, Chris Custer, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, amWhy, Micah, Chris Custer, user10354138 14 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, amWhy, Micah, Chris Custer, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
3
Would you be able to solve $x^3=-8$ instead? If so, try substituting $1+z^2 = x$. If not, this should be your question.
– Florian
yesterday
add a comment |
3
Would you be able to solve $x^3=-8$ instead? If so, try substituting $1+z^2 = x$. If not, this should be your question.
– Florian
yesterday
3
3
Would you be able to solve $x^3=-8$ instead? If so, try substituting $1+z^2 = x$. If not, this should be your question.
– Florian
yesterday
Would you be able to solve $x^3=-8$ instead? If so, try substituting $1+z^2 = x$. If not, this should be your question.
– Florian
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Hint: $-8=8 e^{(pi + 2 pi k)i} $ where $k in mathbb{Z}$. Can you now manipulate the equation algebraically? This will get you all the solutions.
As a start:
$$(z^2 +1)^3=8 e^{(pi + 2 pi k)i} $$
$$z^2 +1=2 e^{frac{1}{3}(pi + 2 pi k)i} $$
Alternatively you can compare the modulus and the arguments of both expressions to figure out what $z$ must be, but a quick hack would now be to write:
$$ z=pmsqrt{2 e^{frac{1}{3}(pi + 2 pi k)i} -1} $$
This should give you all the values for $z$ if you let $k=0,1,2$
answered yesterday
WesleyGroupshaveFeelingsToo
731217
add a comment |
up vote
0
down vote
$$(1+z^2)^3 +8=(z^2+3)(z^4+3)$$
Then solutions is
$$pm sqrt3,i,quad frac{3^{1/4}}{sqrt2}(pm1pm i)$$
answered yesterday
Aleksas Domarkas
6455
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: $-8=8 e^{(pi + 2 pi k)i} $ where $k in mathbb{Z}$. Can you now manipulate the equation algebraically? This will get you all the solutions.
As a start:
$$(z^2 +1)^3=8 e^{(pi + 2 pi k)i} $$
$$z^2 +1=2 e^{frac{1}{3}(pi + 2 pi k)i} $$
Alternatively you can compare the modulus and the arguments of both expressions to figure out what $z$ must be, but a quick hack would now be to write:
$$ z=pmsqrt{2 e^{frac{1}{3}(pi + 2 pi k)i} -1} $$
This should give you all the values for $z$ if you let $k=0,1,2$
answered yesterday
WesleyGroupshaveFeelingsToo
731217
add a comment |
up vote
1
down vote
accepted
Hint: $-8=8 e^{(pi + 2 pi k)i} $ where $k in mathbb{Z}$. Can you now manipulate the equation algebraically? This will get you all the solutions.
As a start:
$$(z^2 +1)^3=8 e^{(pi + 2 pi k)i} $$
$$z^2 +1=2 e^{frac{1}{3}(pi + 2 pi k)i} $$
Alternatively you can compare the modulus and the arguments of both expressions to figure out what $z$ must be, but a quick hack would now be to write:
$$ z=pmsqrt{2 e^{frac{1}{3}(pi + 2 pi k)i} -1} $$
This should give you all the values for $z$ if you let $k=0,1,2$
answered yesterday
WesleyGroupshaveFeelingsToo
731217
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: $-8=8 e^{(pi + 2 pi k)i} $ where $k in mathbb{Z}$. Can you now manipulate the equation algebraically? This will get you all the solutions.
As a start:
$$(z^2 +1)^3=8 e^{(pi + 2 pi k)i} $$
$$z^2 +1=2 e^{frac{1}{3}(pi + 2 pi k)i} $$
Alternatively you can compare the modulus and the arguments of both expressions to figure out what $z$ must be, but a quick hack would now be to write:
$$ z=pmsqrt{2 e^{frac{1}{3}(pi + 2 pi k)i} -1} $$
This should give you all the values for $z$ if you let $k=0,1,2$
answered yesterday
WesleyGroupshaveFeelingsToo
731217
Hint: $-8=8 e^{(pi + 2 pi k)i} $ where $k in mathbb{Z}$. Can you now manipulate the equation algebraically? This will get you all the solutions.
As a start:
$$(z^2 +1)^3=8 e^{(pi + 2 pi k)i} $$
$$z^2 +1=2 e^{frac{1}{3}(pi + 2 pi k)i} $$
Alternatively you can compare the modulus and the arguments of both expressions to figure out what $z$ must be, but a quick hack would now be to write:
$$ z=pmsqrt{2 e^{frac{1}{3}(pi + 2 pi k)i} -1} $$
This should give you all the values for $z$ if you let $k=0,1,2$
answered yesterday
WesleyGroupshaveFeelingsToo
731217
edited yesterday
answered yesterday
WesleyGroupshaveFeelingsToo
731217
answered yesterday
WesleyGroupshaveFeelingsToo
731217
answered yesterday
WesleyGroupshaveFeelingsToo
731217
731217
add a comment |
add a comment |
up vote
0
down vote
$$(1+z^2)^3 +8=(z^2+3)(z^4+3)$$
Then solutions is
$$pm sqrt3,i,quad frac{3^{1/4}}{sqrt2}(pm1pm i)$$
answered yesterday
Aleksas Domarkas
6455
add a comment |
up vote
0
down vote
$$(1+z^2)^3 +8=(z^2+3)(z^4+3)$$
Then solutions is
$$pm sqrt3,i,quad frac{3^{1/4}}{sqrt2}(pm1pm i)$$
answered yesterday
Aleksas Domarkas
6455
add a comment |
up vote
0
down vote
up vote
0
down vote
$$(1+z^2)^3 +8=(z^2+3)(z^4+3)$$
Then solutions is
$$pm sqrt3,i,quad frac{3^{1/4}}{sqrt2}(pm1pm i)$$
answered yesterday
Aleksas Domarkas
6455
$$(1+z^2)^3 +8=(z^2+3)(z^4+3)$$
Then solutions is
$$pm sqrt3,i,quad frac{3^{1/4}}{sqrt2}(pm1pm i)$$
answered yesterday
Aleksas Domarkas
6455
answered yesterday
Aleksas Domarkas
6455
answered yesterday
Aleksas Domarkas
6455
answered yesterday
Aleksas Domarkas
6455
6455
add a comment |
add a comment |
3
Would you be able to solve $x^3=-8$ instead? If so, try substituting $1+z^2 = x$. If not, this should be your question.
– Florian
yesterday