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
87.2k74393
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
87.2k74393
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
87.2k74393
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
87.2k74393
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
87.2k74393
edited Nov 18 at 11:26
answered Nov 18 at 11:19
gimusi
87.2k74393
answered Nov 18 at 11:19
gimusi
87.2k74393
answered Nov 18 at 11:19
gimusi
87.2k74393
87.2k74393
add a comment |
add a comment |
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