Complex roots, conjugated complex numbers
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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
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up vote
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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
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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
90.7k74495
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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
90.7k74495
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
90.7k74495
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
90.7k74495
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
90.7k74495
edited Nov 18 at 11:26
answered Nov 18 at 11:19
gimusi
90.7k74495
answered Nov 18 at 11:19
gimusi
90.7k74495
answered Nov 18 at 11:19
gimusi
90.7k74495
90.7k74495
add a comment |
add a comment |
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Sorry my bad I added cos
– B. Czostek
Nov 18 at 11:21