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How do I find $sqrt[3]-i$?
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I'm asked to evaluate $$sqrt[3]-i$$
I suppose $sqrt[3]-i=(a+bi)$
$$implies (a+bi)^3=-i$$
$$implies Im left( (a+bi)^3 right) =Im left( (-i) right)$$
$$implies 3a^2b-b^3=-1$$
Now how am I supposed to find $a$,$b$?
Aren't there infinitely of them instead of just three?
complex-numbers roots
edited Sep 3 at 7:55
Asaf Karagila♦
294k31410737
asked Sep 3 at 3:40
user588826
564
add a comment |Â
up vote
4
down vote
favorite
I'm asked to evaluate $$sqrt[3]-i$$
I suppose $sqrt[3]-i=(a+bi)$
$$implies (a+bi)^3=-i$$
$$implies Im left( (a+bi)^3 right) =Im left( (-i) right)$$
$$implies 3a^2b-b^3=-1$$
Now how am I supposed to find $a$,$b$?
Aren't there infinitely of them instead of just three?
complex-numbers roots
edited Sep 3 at 7:55
Asaf Karagila♦
294k31410737
asked Sep 3 at 3:40
user588826
564
There cannot be infinitely many of them, because $sqrt[3]-i$ are the roots of the complex polynomial $x^3 = -i$, and the fundamental theorem of algebra, says that this will have exactly three complex roots. Do you know what the polar form / polar coordinates are of a complex number?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:45
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó For every $a$ there is a corresponding $b$.
– user588826
Sep 3 at 3:49
1
But what about $mathcalR( (a+bi)^3 ) = mathcalR(-i)$ ? It gives you another equation, so there is not infinitely many.
– Ahmad Bazzi
Sep 3 at 3:53
You have $3a^2b - b^3 = 1$, but then you also have $a^3 - 3ab^2 = 0$, by equating the real parts. One of these equations may have infinitely many solutions, but now we have two simultaneous equations.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:57
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó Can we deduce something from both of them?
– user588826
Sep 3 at 3:59
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm asked to evaluate $$sqrt[3]-i$$
I suppose $sqrt[3]-i=(a+bi)$
$$implies (a+bi)^3=-i$$
$$implies Im left( (a+bi)^3 right) =Im left( (-i) right)$$
$$implies 3a^2b-b^3=-1$$
Now how am I supposed to find $a$,$b$?
Aren't there infinitely of them instead of just three?
complex-numbers roots
edited Sep 3 at 7:55
Asaf Karagila♦
294k31410737
asked Sep 3 at 3:40
user588826
564
I'm asked to evaluate $$sqrt[3]-i$$
I suppose $sqrt[3]-i=(a+bi)$
$$implies (a+bi)^3=-i$$
$$implies Im left( (a+bi)^3 right) =Im left( (-i) right)$$
$$implies 3a^2b-b^3=-1$$
Now how am I supposed to find $a$,$b$?
Aren't there infinitely of them instead of just three?
complex-numbers roots
edited Sep 3 at 7:55
Asaf Karagila♦
294k31410737
asked Sep 3 at 3:40
user588826
564
edited Sep 3 at 7:55
Asaf Karagila♦
294k31410737
edited Sep 3 at 7:55
Asaf Karagila♦
294k31410737
edited Sep 3 at 7:55
Asaf Karagila♦
294k31410737
294k31410737
asked Sep 3 at 3:40
user588826
564
asked Sep 3 at 3:40
user588826
564
asked Sep 3 at 3:40
user588826
564
564
There cannot be infinitely many of them, because $sqrt[3]-i$ are the roots of the complex polynomial $x^3 = -i$, and the fundamental theorem of algebra, says that this will have exactly three complex roots. Do you know what the polar form / polar coordinates are of a complex number?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:45
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó For every $a$ there is a corresponding $b$.
– user588826
Sep 3 at 3:49
1
But what about $mathcalR( (a+bi)^3 ) = mathcalR(-i)$ ? It gives you another equation, so there is not infinitely many.
– Ahmad Bazzi
Sep 3 at 3:53
You have $3a^2b - b^3 = 1$, but then you also have $a^3 - 3ab^2 = 0$, by equating the real parts. One of these equations may have infinitely many solutions, but now we have two simultaneous equations.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:57
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó Can we deduce something from both of them?
– user588826
Sep 3 at 3:59
add a comment |Â
There cannot be infinitely many of them, because $sqrt[3]-i$ are the roots of the complex polynomial $x^3 = -i$, and the fundamental theorem of algebra, says that this will have exactly three complex roots. Do you know what the polar form / polar coordinates are of a complex number?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:45
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó For every $a$ there is a corresponding $b$.
– user588826
Sep 3 at 3:49
1
But what about $mathcalR( (a+bi)^3 ) = mathcalR(-i)$ ? It gives you another equation, so there is not infinitely many.
– Ahmad Bazzi
Sep 3 at 3:53
You have $3a^2b - b^3 = 1$, but then you also have $a^3 - 3ab^2 = 0$, by equating the real parts. One of these equations may have infinitely many solutions, but now we have two simultaneous equations.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:57
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó Can we deduce something from both of them?
– user588826
Sep 3 at 3:59
There cannot be infinitely many of them, because $sqrt[3]-i$ are the roots of the complex polynomial $x^3 = -i$, and the fundamental theorem of algebra, says that this will have exactly three complex roots. Do you know what the polar form / polar coordinates are of a complex number?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:45
There cannot be infinitely many of them, because $sqrt[3]-i$ are the roots of the complex polynomial $x^3 = -i$, and the fundamental theorem of algebra, says that this will have exactly three complex roots. Do you know what the polar form / polar coordinates are of a complex number?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:45
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó For every $a$ there is a corresponding $b$.
– user588826
Sep 3 at 3:49
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó For every $a$ there is a corresponding $b$.
– user588826
Sep 3 at 3:49
1
1
But what about $mathcalR( (a+bi)^3 ) = mathcalR(-i)$ ? It gives you another equation, so there is not infinitely many.
– Ahmad Bazzi
Sep 3 at 3:53
But what about $mathcalR( (a+bi)^3 ) = mathcalR(-i)$ ? It gives you another equation, so there is not infinitely many.
– Ahmad Bazzi
Sep 3 at 3:53
You have $3a^2b - b^3 = 1$, but then you also have $a^3 - 3ab^2 = 0$, by equating the real parts. One of these equations may have infinitely many solutions, but now we have two simultaneous equations.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:57
You have $3a^2b - b^3 = 1$, but then you also have $a^3 - 3ab^2 = 0$, by equating the real parts. One of these equations may have infinitely many solutions, but now we have two simultaneous equations.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:57
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó Can we deduce something from both of them?
– user588826
Sep 3 at 3:59
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó Can we deduce something from both of them?
– user588826
Sep 3 at 3:59
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
4
down vote
accepted
Let us start as you did:
$sqrt [3] -i= a+bi$
$-i=(a+bi)^3$
$-i=a^3+3a^2bi-3ab^2-b^3i$
Therefore we get $2$ equations:
$a^3-3ab^2=0 ldots(1)$
$3a^2b-b^3=-1 ldots (2)$
Solving for $(1)$:
$a(a^2-3b^2)=0$
$therefore a(a-bsqrt3)(a+bsqrt 3)=0$
So:
$a=0$
$a=bsqrt3$
$a=-bsqrt 3$
Solving for $(2)$:
$3a^2b-b^3=-1$
When $a=0$:
$-b^3=-1$
$b^3=1$
$b=1$
When $a=sqrt 3b^2$:
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=fracsqrt 32$
When $a=-sqrt 3b^2$
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=-fracsqrt 32$
So your $3$ solutions are:
$(a,b)=(0,1);(fracsqrt 32,-frac 12);(-fracsqrt 32,-frac 12)$
I sincerely hope this helps you to understand why there are only $3$ solutions without mentioning the fundamental theorem of algebra. Good luck!
answered Sep 3 at 4:12
Mohammad Zuhair Khan
866422
Thank you very much. This helps!
– user588826
Sep 3 at 4:15
1
I aim to please. However I recommend reading en.wikipedia.org/wiki/Fundamental_theorem_of_algebra for more information.
– Mohammad Zuhair Khan
Sep 3 at 4:18
add a comment |Â
up vote
3
down vote
You can solve this as
beginequation
(-i)^frac13
=
( e^i frac3pi2 + i2kpi)^frac13
endequation
which gives us three distinct roots $e^i z_k$, for $k = 0,1,2$, where
beginalign
z_0 &= frac13frac3pi2 + frac2(0)pi3 = fracpi2 \
z_1 &= frac13frac3pi2 + frac2(1)pi3 = frac7pi6 \
z_2 &= frac13frac3pi2 + frac2(2)pi3 = frac11pi6 \
endalign
If you insist on solving it your way, then
beginequation
(a+bi)^3 = -i
endequation
means
beginequation
a^3 + 3a^2bi - 3ab^2 - b^3i = -i
endequation
which means
beginalign
a^3 - 3ab^2 &= 0 \
3a^2b - b^3 &= -1
endalign
which is
beginalign
a^2 - 3b^2 &= 0 \
3a^2b - b^3 &= -1
endalign
or
beginalign
(a - sqrt3 b)(a + sqrt3 b) &= 0 \
3a^2b - b^3 &= -1
endalign
The first equation suggests either
beginequation
a = pm sqrt3 b
endequation
Replacing this in the second equation gives
beginequation
3(sqrt3 b)^2b - b^3 = -1
endequation
which is
beginequation
9b^3 - b^3 = -1
endequation
i.e.
beginequation
b = -frac12
endequation
This will give us
beginequation
a = pm sqrt3 (-frac12)
endequation
which means we get two solutions
beginalign
(a_1,b_1) &= (-sqrt3,-frac12) \
(a_2,b_2) &= (sqrt3,-frac12) \
endalign
which are actually what we found before, i.e.
beginalign
a_1 + ib_1 &= e^i z_1 \
a_2 + ib_2 &= e^i z_2 \
endalign
Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that
beginalign
a_0 + ib_0 &= e^i z_0
endalign
answered Sep 3 at 4:02
Ahmad Bazzi
4,5201623
add a comment |Â
up vote
2
down vote
Alt. hint: Â the cube roots of $,-i,$ are the solutions to $,z^3=-i iff z^3+i=0,$. Using that $,i = -i^3,$ and the identity $,a^3-b^3=(a-b)(a^2+ab+b^2),$, the latter equation can be written as:
$$
0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)big(-(iz)^2+iz-1big)
$$
The first factor gives the (obvious) solution $,z=i,$, and the second factor is a real quadratic in $,iz,$ with roots $,iz = (1 pm i sqrt3)/2 iff z = ldots$
answered Sep 3 at 6:52
dxiv
55.8k64798
add a comment |Â
up vote
1
down vote
When you take the real part in each side, you get another equation. I don't think that is the best way to tackle the problem though. I think it is better:
$$z^3=-i$$
Then we know the magnitude of $z$ is $1$, and the angle is such that when multiplied by 3 it lies at $frac3pi2+2kpi, kin textZ$ (this is pointing downwards). The possible angles are:
$$fracpi2+frac2kpi3$$
So take the cases $k=0, k=1 $ and $k=2$ and you are done, because then the angles start to loop (just adding $2pi$). The solutions are:
$$z=Cos(fracpi2+frac2kpi3) + iSin(fracpi2+frac2kpi3), kintext0,1,2$$
answered Sep 3 at 4:07
Francisco Abusleme
1423
add a comment |Â
up vote
0
down vote
It is easier to find the cube roots of -i using the polar form.
Note that $-i = e^3ipi /2$ thus the cube roots are $e^ipi/2,e^ipi/2+2ipi/3, e^ipi/2+4ipi/3$
You can easily find these roots in $a+bi $ form if you wish to do so.
answered Sep 3 at 4:00
Mohammad Riazi-Kermani
30.7k41852
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let us start as you did:
$sqrt [3] -i= a+bi$
$-i=(a+bi)^3$
$-i=a^3+3a^2bi-3ab^2-b^3i$
Therefore we get $2$ equations:
$a^3-3ab^2=0 ldots(1)$
$3a^2b-b^3=-1 ldots (2)$
Solving for $(1)$:
$a(a^2-3b^2)=0$
$therefore a(a-bsqrt3)(a+bsqrt 3)=0$
So:
$a=0$
$a=bsqrt3$
$a=-bsqrt 3$
Solving for $(2)$:
$3a^2b-b^3=-1$
When $a=0$:
$-b^3=-1$
$b^3=1$
$b=1$
When $a=sqrt 3b^2$:
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=fracsqrt 32$
When $a=-sqrt 3b^2$
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=-fracsqrt 32$
So your $3$ solutions are:
$(a,b)=(0,1);(fracsqrt 32,-frac 12);(-fracsqrt 32,-frac 12)$
I sincerely hope this helps you to understand why there are only $3$ solutions without mentioning the fundamental theorem of algebra. Good luck!
answered Sep 3 at 4:12
Mohammad Zuhair Khan
866422
Thank you very much. This helps!
– user588826
Sep 3 at 4:15
1
I aim to please. However I recommend reading en.wikipedia.org/wiki/Fundamental_theorem_of_algebra for more information.
– Mohammad Zuhair Khan
Sep 3 at 4:18
add a comment |Â
up vote
4
down vote
accepted
Let us start as you did:
$sqrt [3] -i= a+bi$
$-i=(a+bi)^3$
$-i=a^3+3a^2bi-3ab^2-b^3i$
Therefore we get $2$ equations:
$a^3-3ab^2=0 ldots(1)$
$3a^2b-b^3=-1 ldots (2)$
Solving for $(1)$:
$a(a^2-3b^2)=0$
$therefore a(a-bsqrt3)(a+bsqrt 3)=0$
So:
$a=0$
$a=bsqrt3$
$a=-bsqrt 3$
Solving for $(2)$:
$3a^2b-b^3=-1$
When $a=0$:
$-b^3=-1$
$b^3=1$
$b=1$
When $a=sqrt 3b^2$:
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=fracsqrt 32$
When $a=-sqrt 3b^2$
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=-fracsqrt 32$
So your $3$ solutions are:
$(a,b)=(0,1);(fracsqrt 32,-frac 12);(-fracsqrt 32,-frac 12)$
I sincerely hope this helps you to understand why there are only $3$ solutions without mentioning the fundamental theorem of algebra. Good luck!
answered Sep 3 at 4:12
Mohammad Zuhair Khan
866422
Thank you very much. This helps!
– user588826
Sep 3 at 4:15
1
I aim to please. However I recommend reading en.wikipedia.org/wiki/Fundamental_theorem_of_algebra for more information.
– Mohammad Zuhair Khan
Sep 3 at 4:18
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let us start as you did:
$sqrt [3] -i= a+bi$
$-i=(a+bi)^3$
$-i=a^3+3a^2bi-3ab^2-b^3i$
Therefore we get $2$ equations:
$a^3-3ab^2=0 ldots(1)$
$3a^2b-b^3=-1 ldots (2)$
Solving for $(1)$:
$a(a^2-3b^2)=0$
$therefore a(a-bsqrt3)(a+bsqrt 3)=0$
So:
$a=0$
$a=bsqrt3$
$a=-bsqrt 3$
Solving for $(2)$:
$3a^2b-b^3=-1$
When $a=0$:
$-b^3=-1$
$b^3=1$
$b=1$
When $a=sqrt 3b^2$:
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=fracsqrt 32$
When $a=-sqrt 3b^2$
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=-fracsqrt 32$
So your $3$ solutions are:
$(a,b)=(0,1);(fracsqrt 32,-frac 12);(-fracsqrt 32,-frac 12)$
I sincerely hope this helps you to understand why there are only $3$ solutions without mentioning the fundamental theorem of algebra. Good luck!
answered Sep 3 at 4:12
Mohammad Zuhair Khan
866422
Let us start as you did:
$sqrt [3] -i= a+bi$
$-i=(a+bi)^3$
$-i=a^3+3a^2bi-3ab^2-b^3i$
Therefore we get $2$ equations:
$a^3-3ab^2=0 ldots(1)$
$3a^2b-b^3=-1 ldots (2)$
Solving for $(1)$:
$a(a^2-3b^2)=0$
$therefore a(a-bsqrt3)(a+bsqrt 3)=0$
So:
$a=0$
$a=bsqrt3$
$a=-bsqrt 3$
Solving for $(2)$:
$3a^2b-b^3=-1$
When $a=0$:
$-b^3=-1$
$b^3=1$
$b=1$
When $a=sqrt 3b^2$:
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=fracsqrt 32$
When $a=-sqrt 3b^2$
$9b^3-b^3=-1$
$b^3=-frac 18$
$b=-frac 12$
$therefore a=-fracsqrt 32$
So your $3$ solutions are:
$(a,b)=(0,1);(fracsqrt 32,-frac 12);(-fracsqrt 32,-frac 12)$
I sincerely hope this helps you to understand why there are only $3$ solutions without mentioning the fundamental theorem of algebra. Good luck!
answered Sep 3 at 4:12
Mohammad Zuhair Khan
866422
edited 2 days ago
answered Sep 3 at 4:12
Mohammad Zuhair Khan
866422
answered Sep 3 at 4:12
Mohammad Zuhair Khan
866422
answered Sep 3 at 4:12
Mohammad Zuhair Khan
866422
866422
Thank you very much. This helps!
– user588826
Sep 3 at 4:15
1
I aim to please. However I recommend reading en.wikipedia.org/wiki/Fundamental_theorem_of_algebra for more information.
– Mohammad Zuhair Khan
Sep 3 at 4:18
add a comment |Â
Thank you very much. This helps!
– user588826
Sep 3 at 4:15
1
I aim to please. However I recommend reading en.wikipedia.org/wiki/Fundamental_theorem_of_algebra for more information.
– Mohammad Zuhair Khan
Sep 3 at 4:18
Thank you very much. This helps!
– user588826
Sep 3 at 4:15
Thank you very much. This helps!
– user588826
Sep 3 at 4:15
1
1
I aim to please. However I recommend reading en.wikipedia.org/wiki/Fundamental_theorem_of_algebra for more information.
– Mohammad Zuhair Khan
Sep 3 at 4:18
I aim to please. However I recommend reading en.wikipedia.org/wiki/Fundamental_theorem_of_algebra for more information.
– Mohammad Zuhair Khan
Sep 3 at 4:18
add a comment |Â
up vote
3
down vote
You can solve this as
beginequation
(-i)^frac13
=
( e^i frac3pi2 + i2kpi)^frac13
endequation
which gives us three distinct roots $e^i z_k$, for $k = 0,1,2$, where
beginalign
z_0 &= frac13frac3pi2 + frac2(0)pi3 = fracpi2 \
z_1 &= frac13frac3pi2 + frac2(1)pi3 = frac7pi6 \
z_2 &= frac13frac3pi2 + frac2(2)pi3 = frac11pi6 \
endalign
If you insist on solving it your way, then
beginequation
(a+bi)^3 = -i
endequation
means
beginequation
a^3 + 3a^2bi - 3ab^2 - b^3i = -i
endequation
which means
beginalign
a^3 - 3ab^2 &= 0 \
3a^2b - b^3 &= -1
endalign
which is
beginalign
a^2 - 3b^2 &= 0 \
3a^2b - b^3 &= -1
endalign
or
beginalign
(a - sqrt3 b)(a + sqrt3 b) &= 0 \
3a^2b - b^3 &= -1
endalign
The first equation suggests either
beginequation
a = pm sqrt3 b
endequation
Replacing this in the second equation gives
beginequation
3(sqrt3 b)^2b - b^3 = -1
endequation
which is
beginequation
9b^3 - b^3 = -1
endequation
i.e.
beginequation
b = -frac12
endequation
This will give us
beginequation
a = pm sqrt3 (-frac12)
endequation
which means we get two solutions
beginalign
(a_1,b_1) &= (-sqrt3,-frac12) \
(a_2,b_2) &= (sqrt3,-frac12) \
endalign
which are actually what we found before, i.e.
beginalign
a_1 + ib_1 &= e^i z_1 \
a_2 + ib_2 &= e^i z_2 \
endalign
Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that
beginalign
a_0 + ib_0 &= e^i z_0
endalign
answered Sep 3 at 4:02
Ahmad Bazzi
4,5201623
add a comment |Â
up vote
3
down vote
You can solve this as
beginequation
(-i)^frac13
=
( e^i frac3pi2 + i2kpi)^frac13
endequation
which gives us three distinct roots $e^i z_k$, for $k = 0,1,2$, where
beginalign
z_0 &= frac13frac3pi2 + frac2(0)pi3 = fracpi2 \
z_1 &= frac13frac3pi2 + frac2(1)pi3 = frac7pi6 \
z_2 &= frac13frac3pi2 + frac2(2)pi3 = frac11pi6 \
endalign
If you insist on solving it your way, then
beginequation
(a+bi)^3 = -i
endequation
means
beginequation
a^3 + 3a^2bi - 3ab^2 - b^3i = -i
endequation
which means
beginalign
a^3 - 3ab^2 &= 0 \
3a^2b - b^3 &= -1
endalign
which is
beginalign
a^2 - 3b^2 &= 0 \
3a^2b - b^3 &= -1
endalign
or
beginalign
(a - sqrt3 b)(a + sqrt3 b) &= 0 \
3a^2b - b^3 &= -1
endalign
The first equation suggests either
beginequation
a = pm sqrt3 b
endequation
Replacing this in the second equation gives
beginequation
3(sqrt3 b)^2b - b^3 = -1
endequation
which is
beginequation
9b^3 - b^3 = -1
endequation
i.e.
beginequation
b = -frac12
endequation
This will give us
beginequation
a = pm sqrt3 (-frac12)
endequation
which means we get two solutions
beginalign
(a_1,b_1) &= (-sqrt3,-frac12) \
(a_2,b_2) &= (sqrt3,-frac12) \
endalign
which are actually what we found before, i.e.
beginalign
a_1 + ib_1 &= e^i z_1 \
a_2 + ib_2 &= e^i z_2 \
endalign
Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that
beginalign
a_0 + ib_0 &= e^i z_0
endalign
answered Sep 3 at 4:02
Ahmad Bazzi
4,5201623
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You can solve this as
beginequation
(-i)^frac13
=
( e^i frac3pi2 + i2kpi)^frac13
endequation
which gives us three distinct roots $e^i z_k$, for $k = 0,1,2$, where
beginalign
z_0 &= frac13frac3pi2 + frac2(0)pi3 = fracpi2 \
z_1 &= frac13frac3pi2 + frac2(1)pi3 = frac7pi6 \
z_2 &= frac13frac3pi2 + frac2(2)pi3 = frac11pi6 \
endalign
If you insist on solving it your way, then
beginequation
(a+bi)^3 = -i
endequation
means
beginequation
a^3 + 3a^2bi - 3ab^2 - b^3i = -i
endequation
which means
beginalign
a^3 - 3ab^2 &= 0 \
3a^2b - b^3 &= -1
endalign
which is
beginalign
a^2 - 3b^2 &= 0 \
3a^2b - b^3 &= -1
endalign
or
beginalign
(a - sqrt3 b)(a + sqrt3 b) &= 0 \
3a^2b - b^3 &= -1
endalign
The first equation suggests either
beginequation
a = pm sqrt3 b
endequation
Replacing this in the second equation gives
beginequation
3(sqrt3 b)^2b - b^3 = -1
endequation
which is
beginequation
9b^3 - b^3 = -1
endequation
i.e.
beginequation
b = -frac12
endequation
This will give us
beginequation
a = pm sqrt3 (-frac12)
endequation
which means we get two solutions
beginalign
(a_1,b_1) &= (-sqrt3,-frac12) \
(a_2,b_2) &= (sqrt3,-frac12) \
endalign
which are actually what we found before, i.e.
beginalign
a_1 + ib_1 &= e^i z_1 \
a_2 + ib_2 &= e^i z_2 \
endalign
Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that
beginalign
a_0 + ib_0 &= e^i z_0
endalign
answered Sep 3 at 4:02
Ahmad Bazzi
4,5201623
You can solve this as
beginequation
(-i)^frac13
=
( e^i frac3pi2 + i2kpi)^frac13
endequation
which gives us three distinct roots $e^i z_k$, for $k = 0,1,2$, where
beginalign
z_0 &= frac13frac3pi2 + frac2(0)pi3 = fracpi2 \
z_1 &= frac13frac3pi2 + frac2(1)pi3 = frac7pi6 \
z_2 &= frac13frac3pi2 + frac2(2)pi3 = frac11pi6 \
endalign
If you insist on solving it your way, then
beginequation
(a+bi)^3 = -i
endequation
means
beginequation
a^3 + 3a^2bi - 3ab^2 - b^3i = -i
endequation
which means
beginalign
a^3 - 3ab^2 &= 0 \
3a^2b - b^3 &= -1
endalign
which is
beginalign
a^2 - 3b^2 &= 0 \
3a^2b - b^3 &= -1
endalign
or
beginalign
(a - sqrt3 b)(a + sqrt3 b) &= 0 \
3a^2b - b^3 &= -1
endalign
The first equation suggests either
beginequation
a = pm sqrt3 b
endequation
Replacing this in the second equation gives
beginequation
3(sqrt3 b)^2b - b^3 = -1
endequation
which is
beginequation
9b^3 - b^3 = -1
endequation
i.e.
beginequation
b = -frac12
endequation
This will give us
beginequation
a = pm sqrt3 (-frac12)
endequation
which means we get two solutions
beginalign
(a_1,b_1) &= (-sqrt3,-frac12) \
(a_2,b_2) &= (sqrt3,-frac12) \
endalign
which are actually what we found before, i.e.
beginalign
a_1 + ib_1 &= e^i z_1 \
a_2 + ib_2 &= e^i z_2 \
endalign
Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that
beginalign
a_0 + ib_0 &= e^i z_0
endalign
answered Sep 3 at 4:02
Ahmad Bazzi
4,5201623
edited Sep 3 at 4:14
answered Sep 3 at 4:02
Ahmad Bazzi
4,5201623
answered Sep 3 at 4:02
Ahmad Bazzi
4,5201623
answered Sep 3 at 4:02
Ahmad Bazzi
4,5201623
4,5201623
add a comment |Â
add a comment |Â
up vote
2
down vote
Alt. hint: Â the cube roots of $,-i,$ are the solutions to $,z^3=-i iff z^3+i=0,$. Using that $,i = -i^3,$ and the identity $,a^3-b^3=(a-b)(a^2+ab+b^2),$, the latter equation can be written as:
$$
0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)big(-(iz)^2+iz-1big)
$$
The first factor gives the (obvious) solution $,z=i,$, and the second factor is a real quadratic in $,iz,$ with roots $,iz = (1 pm i sqrt3)/2 iff z = ldots$
answered Sep 3 at 6:52
dxiv
55.8k64798
add a comment |Â
up vote
2
down vote
Alt. hint: Â the cube roots of $,-i,$ are the solutions to $,z^3=-i iff z^3+i=0,$. Using that $,i = -i^3,$ and the identity $,a^3-b^3=(a-b)(a^2+ab+b^2),$, the latter equation can be written as:
$$
0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)big(-(iz)^2+iz-1big)
$$
The first factor gives the (obvious) solution $,z=i,$, and the second factor is a real quadratic in $,iz,$ with roots $,iz = (1 pm i sqrt3)/2 iff z = ldots$
answered Sep 3 at 6:52
dxiv
55.8k64798
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Alt. hint: Â the cube roots of $,-i,$ are the solutions to $,z^3=-i iff z^3+i=0,$. Using that $,i = -i^3,$ and the identity $,a^3-b^3=(a-b)(a^2+ab+b^2),$, the latter equation can be written as:
$$
0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)big(-(iz)^2+iz-1big)
$$
The first factor gives the (obvious) solution $,z=i,$, and the second factor is a real quadratic in $,iz,$ with roots $,iz = (1 pm i sqrt3)/2 iff z = ldots$
answered Sep 3 at 6:52
dxiv
55.8k64798
Alt. hint: Â the cube roots of $,-i,$ are the solutions to $,z^3=-i iff z^3+i=0,$. Using that $,i = -i^3,$ and the identity $,a^3-b^3=(a-b)(a^2+ab+b^2),$, the latter equation can be written as:
$$
0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)big(-(iz)^2+iz-1big)
$$
The first factor gives the (obvious) solution $,z=i,$, and the second factor is a real quadratic in $,iz,$ with roots $,iz = (1 pm i sqrt3)/2 iff z = ldots$
answered Sep 3 at 6:52
dxiv
55.8k64798
answered Sep 3 at 6:52
dxiv
55.8k64798
answered Sep 3 at 6:52
dxiv
55.8k64798
answered Sep 3 at 6:52
dxiv
55.8k64798
55.8k64798
add a comment |Â
add a comment |Â
up vote
1
down vote
When you take the real part in each side, you get another equation. I don't think that is the best way to tackle the problem though. I think it is better:
$$z^3=-i$$
Then we know the magnitude of $z$ is $1$, and the angle is such that when multiplied by 3 it lies at $frac3pi2+2kpi, kin textZ$ (this is pointing downwards). The possible angles are:
$$fracpi2+frac2kpi3$$
So take the cases $k=0, k=1 $ and $k=2$ and you are done, because then the angles start to loop (just adding $2pi$). The solutions are:
$$z=Cos(fracpi2+frac2kpi3) + iSin(fracpi2+frac2kpi3), kintext0,1,2$$
answered Sep 3 at 4:07
Francisco Abusleme
1423
add a comment |Â
up vote
1
down vote
When you take the real part in each side, you get another equation. I don't think that is the best way to tackle the problem though. I think it is better:
$$z^3=-i$$
Then we know the magnitude of $z$ is $1$, and the angle is such that when multiplied by 3 it lies at $frac3pi2+2kpi, kin textZ$ (this is pointing downwards). The possible angles are:
$$fracpi2+frac2kpi3$$
So take the cases $k=0, k=1 $ and $k=2$ and you are done, because then the angles start to loop (just adding $2pi$). The solutions are:
$$z=Cos(fracpi2+frac2kpi3) + iSin(fracpi2+frac2kpi3), kintext0,1,2$$
answered Sep 3 at 4:07
Francisco Abusleme
1423
add a comment |Â
up vote
1
down vote
up vote
1
down vote
When you take the real part in each side, you get another equation. I don't think that is the best way to tackle the problem though. I think it is better:
$$z^3=-i$$
Then we know the magnitude of $z$ is $1$, and the angle is such that when multiplied by 3 it lies at $frac3pi2+2kpi, kin textZ$ (this is pointing downwards). The possible angles are:
$$fracpi2+frac2kpi3$$
So take the cases $k=0, k=1 $ and $k=2$ and you are done, because then the angles start to loop (just adding $2pi$). The solutions are:
$$z=Cos(fracpi2+frac2kpi3) + iSin(fracpi2+frac2kpi3), kintext0,1,2$$
answered Sep 3 at 4:07
Francisco Abusleme
1423
When you take the real part in each side, you get another equation. I don't think that is the best way to tackle the problem though. I think it is better:
$$z^3=-i$$
Then we know the magnitude of $z$ is $1$, and the angle is such that when multiplied by 3 it lies at $frac3pi2+2kpi, kin textZ$ (this is pointing downwards). The possible angles are:
$$fracpi2+frac2kpi3$$
So take the cases $k=0, k=1 $ and $k=2$ and you are done, because then the angles start to loop (just adding $2pi$). The solutions are:
$$z=Cos(fracpi2+frac2kpi3) + iSin(fracpi2+frac2kpi3), kintext0,1,2$$
answered Sep 3 at 4:07
Francisco Abusleme
1423
edited Sep 3 at 22:07
answered Sep 3 at 4:07
Francisco Abusleme
1423
answered Sep 3 at 4:07
Francisco Abusleme
1423
answered Sep 3 at 4:07
Francisco Abusleme
1423
1423
add a comment |Â
add a comment |Â
up vote
0
down vote
It is easier to find the cube roots of -i using the polar form.
Note that $-i = e^3ipi /2$ thus the cube roots are $e^ipi/2,e^ipi/2+2ipi/3, e^ipi/2+4ipi/3$
You can easily find these roots in $a+bi $ form if you wish to do so.
answered Sep 3 at 4:00
Mohammad Riazi-Kermani
30.7k41852
add a comment |Â
up vote
0
down vote
It is easier to find the cube roots of -i using the polar form.
Note that $-i = e^3ipi /2$ thus the cube roots are $e^ipi/2,e^ipi/2+2ipi/3, e^ipi/2+4ipi/3$
You can easily find these roots in $a+bi $ form if you wish to do so.
answered Sep 3 at 4:00
Mohammad Riazi-Kermani
30.7k41852
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is easier to find the cube roots of -i using the polar form.
Note that $-i = e^3ipi /2$ thus the cube roots are $e^ipi/2,e^ipi/2+2ipi/3, e^ipi/2+4ipi/3$
You can easily find these roots in $a+bi $ form if you wish to do so.
answered Sep 3 at 4:00
Mohammad Riazi-Kermani
30.7k41852
It is easier to find the cube roots of -i using the polar form.
Note that $-i = e^3ipi /2$ thus the cube roots are $e^ipi/2,e^ipi/2+2ipi/3, e^ipi/2+4ipi/3$
You can easily find these roots in $a+bi $ form if you wish to do so.
answered Sep 3 at 4:00
Mohammad Riazi-Kermani
30.7k41852
answered Sep 3 at 4:00
Mohammad Riazi-Kermani
30.7k41852
answered Sep 3 at 4:00
Mohammad Riazi-Kermani
30.7k41852
answered Sep 3 at 4:00
Mohammad Riazi-Kermani
30.7k41852
30.7k41852
add a comment |Â
add a comment |Â
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There cannot be infinitely many of them, because $sqrt[3]-i$ are the roots of the complex polynomial $x^3 = -i$, and the fundamental theorem of algebra, says that this will have exactly three complex roots. Do you know what the polar form / polar coordinates are of a complex number?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:45
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó For every $a$ there is a corresponding $b$.
– user588826
Sep 3 at 3:49
1
But what about $mathcalR( (a+bi)^3 ) = mathcalR(-i)$ ? It gives you another equation, so there is not infinitely many.
– Ahmad Bazzi
Sep 3 at 3:53
You have $3a^2b - b^3 = 1$, but then you also have $a^3 - 3ab^2 = 0$, by equating the real parts. One of these equations may have infinitely many solutions, but now we have two simultaneous equations.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 3 at 3:57
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó Can we deduce something from both of them?
– user588826
Sep 3 at 3:59