Complex roots, conjugated complex numbers
up vote
0
down vote
favorite
Knowing that $$ cosfrac{pi}{8}=frac {1}{2}sqrt{2+sqrt{2}},$$
find all roots of these equations:
$2 overline z=z^7$,
$32 overline z=z^7$,
$128 overline z+z^7=0$.
Only those which have solutions different from $z=0$.
complex-numbers
asked Nov 18 at 11:09
B. Czostek
294
add a comment |
up vote
0
down vote
favorite
Knowing that $$ cosfrac{pi}{8}=frac {1}{2}sqrt{2+sqrt{2}},$$
find all roots of these equations:
$2 overline z=z^7$,
$32 overline z=z^7$,
$128 overline z+z^7=0$.
Only those which have solutions different from $z=0$.
complex-numbers
asked Nov 18 at 11:09
B. Czostek
294
Sorry my bad I added cos
– B. Czostek
Nov 18 at 11:21
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Knowing that $$ cosfrac{pi}{8}=frac {1}{2}sqrt{2+sqrt{2}},$$
find all roots of these equations:
$2 overline z=z^7$,
$32 overline z=z^7$,
$128 overline z+z^7=0$.
Only those which have solutions different from $z=0$.
complex-numbers
asked Nov 18 at 11:09
B. Czostek
294
Knowing that $$ cosfrac{pi}{8}=frac {1}{2}sqrt{2+sqrt{2}},$$
find all roots of these equations:
$2 overline z=z^7$,
$32 overline z=z^7$,
$128 overline z+z^7=0$.
Only those which have solutions different from $z=0$.
complex-numbers
complex-numbers
asked Nov 18 at 11:09
B. Czostek
294
asked Nov 18 at 11:09
B. Czostek
294
edited Nov 18 at 11:20
asked Nov 18 at 11:09
B. Czostek
294
asked Nov 18 at 11:09
B. Czostek
294
asked Nov 18 at 11:09
B. Czostek
294
294
Sorry my bad I added cos
– B. Czostek
Nov 18 at 11:21
add a comment |
Sorry my bad I added cos
– B. Czostek
Nov 18 at 11:21
Sorry my bad I added cos
– B. Czostek
Nov 18 at 11:21
Sorry my bad I added cos
– B. Czostek
Nov 18 at 11:21
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
For the first one we have that
$$2overline z=z^7 implies 2overline zz=z^8 implies z^8=2|z|^2implies |z|^6=2 quad z=sqrt[6] 2$$
then we need to solve
$$z^8=2sqrt[3] 2$$
and similarly for the others.
The solution seems not related to $cos frac{pi}8$.
answered Nov 18 at 11:19
gimusi
89.4k74495
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For the first one we have that
$$2overline z=z^7 implies 2overline zz=z^8 implies z^8=2|z|^2implies |z|^6=2 quad z=sqrt[6] 2$$
then we need to solve
$$z^8=2sqrt[3] 2$$
and similarly for the others.
The solution seems not related to $cos frac{pi}8$.
answered Nov 18 at 11:19
gimusi
89.4k74495
add a comment |
up vote
0
down vote
For the first one we have that
$$2overline z=z^7 implies 2overline zz=z^8 implies z^8=2|z|^2implies |z|^6=2 quad z=sqrt[6] 2$$
then we need to solve
$$z^8=2sqrt[3] 2$$
and similarly for the others.
The solution seems not related to $cos frac{pi}8$.
answered Nov 18 at 11:19
gimusi
89.4k74495
add a comment |
up vote
0
down vote
up vote
0
down vote
For the first one we have that
$$2overline z=z^7 implies 2overline zz=z^8 implies z^8=2|z|^2implies |z|^6=2 quad z=sqrt[6] 2$$
then we need to solve
$$z^8=2sqrt[3] 2$$
and similarly for the others.
The solution seems not related to $cos frac{pi}8$.
answered Nov 18 at 11:19
gimusi
89.4k74495
For the first one we have that
$$2overline z=z^7 implies 2overline zz=z^8 implies z^8=2|z|^2implies |z|^6=2 quad z=sqrt[6] 2$$
then we need to solve
$$z^8=2sqrt[3] 2$$
and similarly for the others.
The solution seems not related to $cos frac{pi}8$.
answered Nov 18 at 11:19
gimusi
89.4k74495
edited Nov 18 at 11:26
answered Nov 18 at 11:19
gimusi
89.4k74495
answered Nov 18 at 11:19
gimusi
89.4k74495
answered Nov 18 at 11:19
gimusi
89.4k74495
89.4k74495
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%2f3003404%2fcomplex-roots-conjugated-complex-numbers%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
Sorry my bad I added cos
– B. Czostek
Nov 18 at 11:21