Mixing
Finding roots using Euler's formula
$begingroup$
I'm going through past exam papers and have come accross the following question:
find all solutions to $z^5=2-2i $
for this question I was going to use Euler's formula:
$ e^{i(2k+1)pi}=-1 $
$ -2e^{i(2k+1)pi}=2 $
$ -2e^{i(2k+1)pi}-2i=2-2i $
$ z^5 =-2e^{i(2k+1)pi}-2i$
$ z = -2^{frac 1 5}e^{frac {i(2k+1)pi}{5}}-2i^{frac 1 5}$
I'll then sub in k=0,1,2,3,4
am I on the right track here?
thanks!
complex-analysis complex-numbers
asked Jan 5 '17 at 12:42
user6122081user6122081
758
$endgroup$
add a comment |
$begingroup$
I'm going through past exam papers and have come accross the following question:
find all solutions to $z^5=2-2i $
for this question I was going to use Euler's formula:
$ e^{i(2k+1)pi}=-1 $
$ -2e^{i(2k+1)pi}=2 $
$ -2e^{i(2k+1)pi}-2i=2-2i $
$ z^5 =-2e^{i(2k+1)pi}-2i$
$ z = -2^{frac 1 5}e^{frac {i(2k+1)pi}{5}}-2i^{frac 1 5}$
I'll then sub in k=0,1,2,3,4
am I on the right track here?
thanks!
complex-analysis complex-numbers
asked Jan 5 '17 at 12:42
user6122081user6122081
758
$endgroup$
$begingroup$
Try writing $2-2i$ in the $e^{x}$ form first
$endgroup$
– TheMathsGeek
Jan 5 '17 at 12:47
2
$begingroup$
$(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
$endgroup$
– ajotatxe
Jan 5 '17 at 12:47
add a comment |
$begingroup$
I'm going through past exam papers and have come accross the following question:
find all solutions to $z^5=2-2i $
for this question I was going to use Euler's formula:
$ e^{i(2k+1)pi}=-1 $
$ -2e^{i(2k+1)pi}=2 $
$ -2e^{i(2k+1)pi}-2i=2-2i $
$ z^5 =-2e^{i(2k+1)pi}-2i$
$ z = -2^{frac 1 5}e^{frac {i(2k+1)pi}{5}}-2i^{frac 1 5}$
I'll then sub in k=0,1,2,3,4
am I on the right track here?
thanks!
complex-analysis complex-numbers
asked Jan 5 '17 at 12:42
user6122081user6122081
758
$endgroup$
I'm going through past exam papers and have come accross the following question:
find all solutions to $z^5=2-2i $
for this question I was going to use Euler's formula:
$ e^{i(2k+1)pi}=-1 $
$ -2e^{i(2k+1)pi}=2 $
$ -2e^{i(2k+1)pi}-2i=2-2i $
$ z^5 =-2e^{i(2k+1)pi}-2i$
$ z = -2^{frac 1 5}e^{frac {i(2k+1)pi}{5}}-2i^{frac 1 5}$
I'll then sub in k=0,1,2,3,4
am I on the right track here?
thanks!
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Jan 5 '17 at 12:42
user6122081user6122081
758
asked Jan 5 '17 at 12:42
user6122081user6122081
758
asked Jan 5 '17 at 12:42
user6122081user6122081
758
asked Jan 5 '17 at 12:42
user6122081user6122081
758
asked Jan 5 '17 at 12:42
user6122081user6122081
758
758
$begingroup$
Try writing $2-2i$ in the $e^{x}$ form first
$endgroup$
– TheMathsGeek
Jan 5 '17 at 12:47
2
$begingroup$
$(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
$endgroup$
– ajotatxe
Jan 5 '17 at 12:47
add a comment |
$begingroup$
Try writing $2-2i$ in the $e^{x}$ form first
$endgroup$
– TheMathsGeek
Jan 5 '17 at 12:47
2
$begingroup$
$(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
$endgroup$
– ajotatxe
Jan 5 '17 at 12:47
$begingroup$
Try writing $2-2i$ in the $e^{x}$ form first
$endgroup$
– TheMathsGeek
Jan 5 '17 at 12:47
$begingroup$
Try writing $2-2i$ in the $e^{x}$ form first
$endgroup$
– TheMathsGeek
Jan 5 '17 at 12:47
2
2
$begingroup$
$(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
$endgroup$
– ajotatxe
Jan 5 '17 at 12:47
$begingroup$
$(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
$endgroup$
– ajotatxe
Jan 5 '17 at 12:47
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint:
$$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$
Therefore
$$2-2i=2sqrt 2e^{7pi/4}$$
answered Jan 5 '17 at 12:50
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
$begingroup$
Using the Euler's Formula is a good approach but I would use:
$$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$
$$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Now compare both:
$$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:
begin{cases}
r^5=2sqrt{2}\
e^{5xi}=e^{(7pi/4+2kpi)i}
end{cases}
P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.
Can you finish?
answered Jan 5 '17 at 12:49
ArnaldoArnaldo
18.1k42246
$endgroup$
$begingroup$
just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
$endgroup$
– user6122081
Jan 5 '17 at 12:54
$begingroup$
I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
$endgroup$
– Arnaldo
Jan 5 '17 at 12:56
$begingroup$
@user6122081: are you ok now?
$endgroup$
– Arnaldo
Jan 5 '17 at 21:11
add a comment |
$begingroup$
We can write $2-2i$ as:
$z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$
now we take the root:
$z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$
with $kin mathbb Z$, $k=0,1,2,3,4$.
Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$
answered Jan 5 '17 at 12:49
MattG88MattG88
2,4682815
$endgroup$
$begingroup$
$arg(2-2i)$ is $displaystyle7frac{pi}{4}$
$endgroup$
– Nosrati
Jan 5 '17 at 14:12
add a comment |
$begingroup$
You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.
So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).
Now $z^5=2sqrt2e^{frac{7pi}4i}$.
So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.
There are $4$ more.
Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.
That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.
answered Jan 2 at 8:57
Chris CusterChris Custer
11.3k3824
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$
Therefore
$$2-2i=2sqrt 2e^{7pi/4}$$
answered Jan 5 '17 at 12:50
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
$begingroup$
Hint:
$$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$
Therefore
$$2-2i=2sqrt 2e^{7pi/4}$$
answered Jan 5 '17 at 12:50
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
$begingroup$
Hint:
$$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$
Therefore
$$2-2i=2sqrt 2e^{7pi/4}$$
answered Jan 5 '17 at 12:50
ajotatxeajotatxe
53.7k23890
$endgroup$
Hint:
$$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$
Therefore
$$2-2i=2sqrt 2e^{7pi/4}$$
answered Jan 5 '17 at 12:50
ajotatxeajotatxe
53.7k23890
answered Jan 5 '17 at 12:50
ajotatxeajotatxe
53.7k23890
answered Jan 5 '17 at 12:50
ajotatxeajotatxe
53.7k23890
answered Jan 5 '17 at 12:50
ajotatxeajotatxe
53.7k23890
53.7k23890
add a comment |
add a comment |
$begingroup$
Using the Euler's Formula is a good approach but I would use:
$$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$
$$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Now compare both:
$$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:
begin{cases}
r^5=2sqrt{2}\
e^{5xi}=e^{(7pi/4+2kpi)i}
end{cases}
P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.
Can you finish?
answered Jan 5 '17 at 12:49
ArnaldoArnaldo
18.1k42246
$endgroup$
$begingroup$
just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
$endgroup$
– user6122081
Jan 5 '17 at 12:54
$begingroup$
I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
$endgroup$
– Arnaldo
Jan 5 '17 at 12:56
$begingroup$
@user6122081: are you ok now?
$endgroup$
– Arnaldo
Jan 5 '17 at 21:11
add a comment |
$begingroup$
Using the Euler's Formula is a good approach but I would use:
$$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$
$$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Now compare both:
$$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:
begin{cases}
r^5=2sqrt{2}\
e^{5xi}=e^{(7pi/4+2kpi)i}
end{cases}
P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.
Can you finish?
answered Jan 5 '17 at 12:49
ArnaldoArnaldo
18.1k42246
$endgroup$
$begingroup$
just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
$endgroup$
– user6122081
Jan 5 '17 at 12:54
$begingroup$
I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
$endgroup$
– Arnaldo
Jan 5 '17 at 12:56
$begingroup$
@user6122081: are you ok now?
$endgroup$
– Arnaldo
Jan 5 '17 at 21:11
add a comment |
$begingroup$
Using the Euler's Formula is a good approach but I would use:
$$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$
$$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Now compare both:
$$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:
begin{cases}
r^5=2sqrt{2}\
e^{5xi}=e^{(7pi/4+2kpi)i}
end{cases}
P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.
Can you finish?
answered Jan 5 '17 at 12:49
ArnaldoArnaldo
18.1k42246
$endgroup$
Using the Euler's Formula is a good approach but I would use:
$$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$
$$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Now compare both:
$$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$
Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:
begin{cases}
r^5=2sqrt{2}\
e^{5xi}=e^{(7pi/4+2kpi)i}
end{cases}
P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.
Can you finish?
answered Jan 5 '17 at 12:49
ArnaldoArnaldo
18.1k42246
edited Jan 5 '17 at 13:07
answered Jan 5 '17 at 12:49
ArnaldoArnaldo
18.1k42246
answered Jan 5 '17 at 12:49
ArnaldoArnaldo
18.1k42246
answered Jan 5 '17 at 12:49
ArnaldoArnaldo
18.1k42246
18.1k42246
$begingroup$
just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
$endgroup$
– user6122081
Jan 5 '17 at 12:54
$begingroup$
I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
$endgroup$
– Arnaldo
Jan 5 '17 at 12:56
$begingroup$
@user6122081: are you ok now?
$endgroup$
– Arnaldo
Jan 5 '17 at 21:11
add a comment |
$begingroup$
just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
$endgroup$
– user6122081
Jan 5 '17 at 12:54
$begingroup$
I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
$endgroup$
– Arnaldo
Jan 5 '17 at 12:56
$begingroup$
@user6122081: are you ok now?
$endgroup$
– Arnaldo
Jan 5 '17 at 21:11
$begingroup$
just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
$endgroup$
– user6122081
Jan 5 '17 at 12:54
$begingroup$
just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
$endgroup$
– user6122081
Jan 5 '17 at 12:54
$begingroup$
I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
$endgroup$
– Arnaldo
Jan 5 '17 at 12:56
$begingroup$
I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
$endgroup$
– Arnaldo
Jan 5 '17 at 12:56
$begingroup$
@user6122081: are you ok now?
$endgroup$
– Arnaldo
Jan 5 '17 at 21:11
$begingroup$
@user6122081: are you ok now?
$endgroup$
– Arnaldo
Jan 5 '17 at 21:11
add a comment |
$begingroup$
We can write $2-2i$ as:
$z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$
now we take the root:
$z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$
with $kin mathbb Z$, $k=0,1,2,3,4$.
Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$
answered Jan 5 '17 at 12:49
MattG88MattG88
2,4682815
$endgroup$
$begingroup$
$arg(2-2i)$ is $displaystyle7frac{pi}{4}$
$endgroup$
– Nosrati
Jan 5 '17 at 14:12
add a comment |
$begingroup$
We can write $2-2i$ as:
$z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$
now we take the root:
$z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$
with $kin mathbb Z$, $k=0,1,2,3,4$.
Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$
answered Jan 5 '17 at 12:49
MattG88MattG88
2,4682815
$endgroup$
$begingroup$
$arg(2-2i)$ is $displaystyle7frac{pi}{4}$
$endgroup$
– Nosrati
Jan 5 '17 at 14:12
add a comment |
$begingroup$
We can write $2-2i$ as:
$z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$
now we take the root:
$z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$
with $kin mathbb Z$, $k=0,1,2,3,4$.
Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$
answered Jan 5 '17 at 12:49
MattG88MattG88
2,4682815
$endgroup$
We can write $2-2i$ as:
$z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$
now we take the root:
$z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$
with $kin mathbb Z$, $k=0,1,2,3,4$.
Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$
answered Jan 5 '17 at 12:49
MattG88MattG88
2,4682815
edited Jan 5 '17 at 17:53
answered Jan 5 '17 at 12:49
MattG88MattG88
2,4682815
answered Jan 5 '17 at 12:49
MattG88MattG88
2,4682815
answered Jan 5 '17 at 12:49
MattG88MattG88
2,4682815
2,4682815
$begingroup$
$arg(2-2i)$ is $displaystyle7frac{pi}{4}$
$endgroup$
– Nosrati
Jan 5 '17 at 14:12
add a comment |
$begingroup$
$arg(2-2i)$ is $displaystyle7frac{pi}{4}$
$endgroup$
– Nosrati
Jan 5 '17 at 14:12
$begingroup$
$arg(2-2i)$ is $displaystyle7frac{pi}{4}$
$endgroup$
– Nosrati
Jan 5 '17 at 14:12
$begingroup$
$arg(2-2i)$ is $displaystyle7frac{pi}{4}$
$endgroup$
– Nosrati
Jan 5 '17 at 14:12
add a comment |
$begingroup$
You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.
So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).
Now $z^5=2sqrt2e^{frac{7pi}4i}$.
So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.
There are $4$ more.
Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.
That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.
answered Jan 2 at 8:57
Chris CusterChris Custer
11.3k3824
$endgroup$
add a comment |
$begingroup$
You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.
So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).
Now $z^5=2sqrt2e^{frac{7pi}4i}$.
So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.
There are $4$ more.
Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.
That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.
answered Jan 2 at 8:57
Chris CusterChris Custer
11.3k3824
$endgroup$
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$begingroup$
You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.
So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).
Now $z^5=2sqrt2e^{frac{7pi}4i}$.
So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.
There are $4$ more.
Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.
That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.
answered Jan 2 at 8:57
Chris CusterChris Custer
11.3k3824
$endgroup$
You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.
So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).
Now $z^5=2sqrt2e^{frac{7pi}4i}$.
So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.
There are $4$ more.
Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.
That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.
answered Jan 2 at 8:57
Chris CusterChris Custer
11.3k3824
edited Jan 2 at 17:43
answered Jan 2 at 8:57
Chris CusterChris Custer
11.3k3824
answered Jan 2 at 8:57
Chris CusterChris Custer
11.3k3824
answered Jan 2 at 8:57
Chris CusterChris Custer
11.3k3824
11.3k3824
add a comment |
add a comment |
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$begingroup$
Try writing $2-2i$ in the $e^{x}$ form first
$endgroup$
– TheMathsGeek
Jan 5 '17 at 12:47
2
$begingroup$
$(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
$endgroup$
– ajotatxe
Jan 5 '17 at 12:47