Mixing
Help with complex numbers
$begingroup$
Algebraic form of the $z^3 = (3 + i)^6$. Can someone help me to solve this? My answer is $z = 8 + 6i$, but I'm not sure that this is true.
trigonometry complex-numbers
edited Jan 20 at 1:10
bjcolby15
1,47611016
asked Jan 19 at 23:19
Aliaksei KlimovichAliaksei Klimovich
516
$endgroup$
add a comment |
$begingroup$
Algebraic form of the $z^3 = (3 + i)^6$. Can someone help me to solve this? My answer is $z = 8 + 6i$, but I'm not sure that this is true.
trigonometry complex-numbers
edited Jan 20 at 1:10
bjcolby15
1,47611016
asked Jan 19 at 23:19
Aliaksei KlimovichAliaksei Klimovich
516
$endgroup$
$begingroup$
Note that $z^3 = left(3 + iright)^6$, by moving the right side value to the left, basically forms a cubic equation in $z$. Thus, there's always $3$ roots, although some may be repeated. Have you accounted for this?
$endgroup$
– John Omielan
Jan 19 at 23:22
$begingroup$
As already mentioned there are three different candidates for $z$. One of them is found correctly.
$endgroup$
– user
Jan 19 at 23:26
add a comment |
$begingroup$
Algebraic form of the $z^3 = (3 + i)^6$. Can someone help me to solve this? My answer is $z = 8 + 6i$, but I'm not sure that this is true.
trigonometry complex-numbers
edited Jan 20 at 1:10
bjcolby15
1,47611016
asked Jan 19 at 23:19
Aliaksei KlimovichAliaksei Klimovich
516
$endgroup$
Algebraic form of the $z^3 = (3 + i)^6$. Can someone help me to solve this? My answer is $z = 8 + 6i$, but I'm not sure that this is true.
trigonometry complex-numbers
trigonometry complex-numbers
edited Jan 20 at 1:10
bjcolby15
1,47611016
asked Jan 19 at 23:19
Aliaksei KlimovichAliaksei Klimovich
516
edited Jan 20 at 1:10
bjcolby15
1,47611016
asked Jan 19 at 23:19
Aliaksei KlimovichAliaksei Klimovich
516
edited Jan 20 at 1:10
bjcolby15
1,47611016
edited Jan 20 at 1:10
bjcolby15
1,47611016
edited Jan 20 at 1:10
bjcolby15
1,47611016
1,47611016
asked Jan 19 at 23:19
Aliaksei KlimovichAliaksei Klimovich
516
asked Jan 19 at 23:19
Aliaksei KlimovichAliaksei Klimovich
516
asked Jan 19 at 23:19
Aliaksei KlimovichAliaksei Klimovich
516
516
$begingroup$
Note that $z^3 = left(3 + iright)^6$, by moving the right side value to the left, basically forms a cubic equation in $z$. Thus, there's always $3$ roots, although some may be repeated. Have you accounted for this?
$endgroup$
– John Omielan
Jan 19 at 23:22
$begingroup$
As already mentioned there are three different candidates for $z$. One of them is found correctly.
$endgroup$
– user
Jan 19 at 23:26
add a comment |
$begingroup$
Note that $z^3 = left(3 + iright)^6$, by moving the right side value to the left, basically forms a cubic equation in $z$. Thus, there's always $3$ roots, although some may be repeated. Have you accounted for this?
$endgroup$
– John Omielan
Jan 19 at 23:22
$begingroup$
As already mentioned there are three different candidates for $z$. One of them is found correctly.
$endgroup$
– user
Jan 19 at 23:26
$begingroup$
Note that $z^3 = left(3 + iright)^6$, by moving the right side value to the left, basically forms a cubic equation in $z$. Thus, there's always $3$ roots, although some may be repeated. Have you accounted for this?
$endgroup$
– John Omielan
Jan 19 at 23:22
$begingroup$
Note that $z^3 = left(3 + iright)^6$, by moving the right side value to the left, basically forms a cubic equation in $z$. Thus, there's always $3$ roots, although some may be repeated. Have you accounted for this?
$endgroup$
– John Omielan
Jan 19 at 23:22
$begingroup$
As already mentioned there are three different candidates for $z$. One of them is found correctly.
$endgroup$
– user
Jan 19 at 23:26
$begingroup$
As already mentioned there are three different candidates for $z$. One of them is found correctly.
$endgroup$
– user
Jan 19 at 23:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let
$$left(3 + iright)^2 = 8 + 6i = a tag{1}label{eq1}$$
so
$$z^3 = left(3 + iright)^6 = left[left(3 + iright)^2right]^3 tag{2}label{eq2}$$
becomes, after moving the right side to the left & factoring, that
$$z^3 - a^3 = left(z - aright) left(z^2 + za + a^2right) = 0 tag{3}label{eq3}$$
Thus, either $z - a = 0$ or $z^2 + za + a^2 = 0$. For the first one $z = a$ becomes, from eqref{eq1}, that $z = 8 + 6i$, as you have already determined. For the second one, you can just use the quadratic formula to get the other $2$ roots as
$$z = cfrac{-a pm sqrt{-3a^2}}{2} = cfrac{-a pm sqrt{-3}a}{2} tag{4}label{eq4}$$
Note this becomes
$$z = left(-cfrac{1}{2} pm cfrac{sqrt{3}}{2}iright)a tag{5}label{eq5}$$
as orion has previously stated in their answer.
answered Jan 19 at 23:30
John OmielanJohn Omielan
3,4801215
$endgroup$
add a comment |
$begingroup$
You found one solution. The other two can be found by rotating the known solution in the complex plane by $pm 120^circ$ ($n$-th roots always form a regular $n$-sided polygon). For rotation, you can multiply by $cosfrac{2pi}{3}pm isinfrac{2pi}{3}=-frac12pm ifrac{sqrt{3}}{2}$.
answered Jan 19 at 23:41
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
Whenever you have $z^3 = k$ there will be $3$ solutions.
If $z^3 = (3+i)^6 = ((3+i)^2)^3$ then one solution is obviously $z = (3+i)^2 = 9 + 6i + i^2 = 9+6i -1 = 8+6i$.
But that is only one solution.
However $w^3 = 1$ has three solutions.
If we let $w$ be one of the solutions and we let $z = w(8+6i) = w(3+i)^2$ then $z^3 = w^3((3+i)^2)^3 = 1*(3+i)^6 = (3+i)^6$.
So if the three solutions to $w^3=1$ are $1, b,c$ then the three solutions to $z^3 = (3+i)^6$ are $8 + 6i, b(8+6i), c(8+6i)$.
So we nee to solve $w^3 = 1$.
Let $w = m+ni$ (where $m,n$ are real) then $(m+ni)^3 = m^3 + 3m^2ni + 3mn^2i^2 + n^3i^3=$
$m^3 + 3m^2ni -3mn^2 - n^3i = $
$(m^3 - 3mn^2) + (3m^2n-n^3)i = $
$1 + 0i$
So $m^3 -3mn^2 =1$ and $3m^2n -n^3=0$.
$3m^2n -n^3= n(3m^2-n^2) = n(msqrt 3 -n)(msqrt 3 + n) = 0$ so
either:
1) $n = 0$ and $m^3 = 1$ so $m = 1$ and $w = 1 + 0i = 1$.
2) $msqrt 3 - n = 0$ and $msqrt 3 = n$ so $m^3 -3mn^2 = 1$ and so
$m^3 - 3m(msqrt 3)^2 = m^3 - 3m(3m^2) =m^3 -9m^3 = -8m^3 =1$ so $m = -frac 12$ and $w = -frac 12 -frac {sqrt 3}2 i$.
3) $msqrt 3+ n = 0$ so $n = -msqrt 3$ and so
$m^3 - 3m(-msqrt 3)^2=.... = 1$ so $m = -frac 12$ and $w = -frac 12 + frac {sqrt 3}2i$.
So the three solutions to $z^3 =(3+i)^6$ are:
1)$8+6i$
2)$(-frac 12 -frac {sqrt 3}2 i)(8 + 6i) = -4 -4sqrt 3i - 3i -3sqrt 3 i^2 = (-4+3sqrt 3) -(4sqrt 3-3)i$.
and 3) $(-frac 12 +frac {sqrt 3}2 i)(8 + 6i) = -4 +4sqrt 3i - 3i +3sqrt 3 i^2 = (-4-3sqrt 3) +(4sqrt 3-3)i$
....
However once you learn polar notation that $z = m + ni = r(cos theta + isin theta) := re^{itheta}$ where $r = sqrt{m^2 + n^2}$ and $theta$ is the solutions to $cos theta = frac mr$ and $sin theta = frac nr$, once you learn that these problems become MUCH easier.
answered Jan 19 at 23:58
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
Let
$$left(3 + iright)^2 = 8 + 6i = a tag{1}label{eq1}$$
so
$$z^3 = left(3 + iright)^6 = left[left(3 + iright)^2right]^3 tag{2}label{eq2}$$
becomes, after moving the right side to the left & factoring, that
$$z^3 - a^3 = left(z - aright) left(z^2 + za + a^2right) = 0 tag{3}label{eq3}$$
Thus, either $z - a = 0$ or $z^2 + za + a^2 = 0$. For the first one $z = a$ becomes, from eqref{eq1}, that $z = 8 + 6i$, as you have already determined. For the second one, you can just use the quadratic formula to get the other $2$ roots as
$$z = cfrac{-a pm sqrt{-3a^2}}{2} = cfrac{-a pm sqrt{-3}a}{2} tag{4}label{eq4}$$
Note this becomes
$$z = left(-cfrac{1}{2} pm cfrac{sqrt{3}}{2}iright)a tag{5}label{eq5}$$
as orion has previously stated in their answer.
answered Jan 19 at 23:30
John OmielanJohn Omielan
3,4801215
$endgroup$
add a comment |
$begingroup$
Let
$$left(3 + iright)^2 = 8 + 6i = a tag{1}label{eq1}$$
so
$$z^3 = left(3 + iright)^6 = left[left(3 + iright)^2right]^3 tag{2}label{eq2}$$
becomes, after moving the right side to the left & factoring, that
$$z^3 - a^3 = left(z - aright) left(z^2 + za + a^2right) = 0 tag{3}label{eq3}$$
Thus, either $z - a = 0$ or $z^2 + za + a^2 = 0$. For the first one $z = a$ becomes, from eqref{eq1}, that $z = 8 + 6i$, as you have already determined. For the second one, you can just use the quadratic formula to get the other $2$ roots as
$$z = cfrac{-a pm sqrt{-3a^2}}{2} = cfrac{-a pm sqrt{-3}a}{2} tag{4}label{eq4}$$
Note this becomes
$$z = left(-cfrac{1}{2} pm cfrac{sqrt{3}}{2}iright)a tag{5}label{eq5}$$
as orion has previously stated in their answer.
answered Jan 19 at 23:30
John OmielanJohn Omielan
3,4801215
$endgroup$
add a comment |
$begingroup$
Let
$$left(3 + iright)^2 = 8 + 6i = a tag{1}label{eq1}$$
so
$$z^3 = left(3 + iright)^6 = left[left(3 + iright)^2right]^3 tag{2}label{eq2}$$
becomes, after moving the right side to the left & factoring, that
$$z^3 - a^3 = left(z - aright) left(z^2 + za + a^2right) = 0 tag{3}label{eq3}$$
Thus, either $z - a = 0$ or $z^2 + za + a^2 = 0$. For the first one $z = a$ becomes, from eqref{eq1}, that $z = 8 + 6i$, as you have already determined. For the second one, you can just use the quadratic formula to get the other $2$ roots as
$$z = cfrac{-a pm sqrt{-3a^2}}{2} = cfrac{-a pm sqrt{-3}a}{2} tag{4}label{eq4}$$
Note this becomes
$$z = left(-cfrac{1}{2} pm cfrac{sqrt{3}}{2}iright)a tag{5}label{eq5}$$
as orion has previously stated in their answer.
answered Jan 19 at 23:30
John OmielanJohn Omielan
3,4801215
$endgroup$
Let
$$left(3 + iright)^2 = 8 + 6i = a tag{1}label{eq1}$$
so
$$z^3 = left(3 + iright)^6 = left[left(3 + iright)^2right]^3 tag{2}label{eq2}$$
becomes, after moving the right side to the left & factoring, that
$$z^3 - a^3 = left(z - aright) left(z^2 + za + a^2right) = 0 tag{3}label{eq3}$$
Thus, either $z - a = 0$ or $z^2 + za + a^2 = 0$. For the first one $z = a$ becomes, from eqref{eq1}, that $z = 8 + 6i$, as you have already determined. For the second one, you can just use the quadratic formula to get the other $2$ roots as
$$z = cfrac{-a pm sqrt{-3a^2}}{2} = cfrac{-a pm sqrt{-3}a}{2} tag{4}label{eq4}$$
Note this becomes
$$z = left(-cfrac{1}{2} pm cfrac{sqrt{3}}{2}iright)a tag{5}label{eq5}$$
as orion has previously stated in their answer.
answered Jan 19 at 23:30
John OmielanJohn Omielan
3,4801215
edited Jan 20 at 1:37
answered Jan 19 at 23:30
John OmielanJohn Omielan
3,4801215
answered Jan 19 at 23:30
John OmielanJohn Omielan
3,4801215
answered Jan 19 at 23:30
John OmielanJohn Omielan
3,4801215
3,4801215
add a comment |
add a comment |
$begingroup$
You found one solution. The other two can be found by rotating the known solution in the complex plane by $pm 120^circ$ ($n$-th roots always form a regular $n$-sided polygon). For rotation, you can multiply by $cosfrac{2pi}{3}pm isinfrac{2pi}{3}=-frac12pm ifrac{sqrt{3}}{2}$.
answered Jan 19 at 23:41
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
You found one solution. The other two can be found by rotating the known solution in the complex plane by $pm 120^circ$ ($n$-th roots always form a regular $n$-sided polygon). For rotation, you can multiply by $cosfrac{2pi}{3}pm isinfrac{2pi}{3}=-frac12pm ifrac{sqrt{3}}{2}$.
answered Jan 19 at 23:41
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
You found one solution. The other two can be found by rotating the known solution in the complex plane by $pm 120^circ$ ($n$-th roots always form a regular $n$-sided polygon). For rotation, you can multiply by $cosfrac{2pi}{3}pm isinfrac{2pi}{3}=-frac12pm ifrac{sqrt{3}}{2}$.
answered Jan 19 at 23:41
orionorion
13.6k11837
$endgroup$
You found one solution. The other two can be found by rotating the known solution in the complex plane by $pm 120^circ$ ($n$-th roots always form a regular $n$-sided polygon). For rotation, you can multiply by $cosfrac{2pi}{3}pm isinfrac{2pi}{3}=-frac12pm ifrac{sqrt{3}}{2}$.
answered Jan 19 at 23:41
orionorion
13.6k11837
answered Jan 19 at 23:41
orionorion
13.6k11837
answered Jan 19 at 23:41
orionorion
13.6k11837
answered Jan 19 at 23:41
orionorion
13.6k11837
13.6k11837
add a comment |
add a comment |
$begingroup$
Whenever you have $z^3 = k$ there will be $3$ solutions.
If $z^3 = (3+i)^6 = ((3+i)^2)^3$ then one solution is obviously $z = (3+i)^2 = 9 + 6i + i^2 = 9+6i -1 = 8+6i$.
But that is only one solution.
However $w^3 = 1$ has three solutions.
If we let $w$ be one of the solutions and we let $z = w(8+6i) = w(3+i)^2$ then $z^3 = w^3((3+i)^2)^3 = 1*(3+i)^6 = (3+i)^6$.
So if the three solutions to $w^3=1$ are $1, b,c$ then the three solutions to $z^3 = (3+i)^6$ are $8 + 6i, b(8+6i), c(8+6i)$.
So we nee to solve $w^3 = 1$.
Let $w = m+ni$ (where $m,n$ are real) then $(m+ni)^3 = m^3 + 3m^2ni + 3mn^2i^2 + n^3i^3=$
$m^3 + 3m^2ni -3mn^2 - n^3i = $
$(m^3 - 3mn^2) + (3m^2n-n^3)i = $
$1 + 0i$
So $m^3 -3mn^2 =1$ and $3m^2n -n^3=0$.
$3m^2n -n^3= n(3m^2-n^2) = n(msqrt 3 -n)(msqrt 3 + n) = 0$ so
either:
1) $n = 0$ and $m^3 = 1$ so $m = 1$ and $w = 1 + 0i = 1$.
2) $msqrt 3 - n = 0$ and $msqrt 3 = n$ so $m^3 -3mn^2 = 1$ and so
$m^3 - 3m(msqrt 3)^2 = m^3 - 3m(3m^2) =m^3 -9m^3 = -8m^3 =1$ so $m = -frac 12$ and $w = -frac 12 -frac {sqrt 3}2 i$.
3) $msqrt 3+ n = 0$ so $n = -msqrt 3$ and so
$m^3 - 3m(-msqrt 3)^2=.... = 1$ so $m = -frac 12$ and $w = -frac 12 + frac {sqrt 3}2i$.
So the three solutions to $z^3 =(3+i)^6$ are:
1)$8+6i$
2)$(-frac 12 -frac {sqrt 3}2 i)(8 + 6i) = -4 -4sqrt 3i - 3i -3sqrt 3 i^2 = (-4+3sqrt 3) -(4sqrt 3-3)i$.
and 3) $(-frac 12 +frac {sqrt 3}2 i)(8 + 6i) = -4 +4sqrt 3i - 3i +3sqrt 3 i^2 = (-4-3sqrt 3) +(4sqrt 3-3)i$
....
However once you learn polar notation that $z = m + ni = r(cos theta + isin theta) := re^{itheta}$ where $r = sqrt{m^2 + n^2}$ and $theta$ is the solutions to $cos theta = frac mr$ and $sin theta = frac nr$, once you learn that these problems become MUCH easier.
answered Jan 19 at 23:58
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
Whenever you have $z^3 = k$ there will be $3$ solutions.
If $z^3 = (3+i)^6 = ((3+i)^2)^3$ then one solution is obviously $z = (3+i)^2 = 9 + 6i + i^2 = 9+6i -1 = 8+6i$.
But that is only one solution.
However $w^3 = 1$ has three solutions.
If we let $w$ be one of the solutions and we let $z = w(8+6i) = w(3+i)^2$ then $z^3 = w^3((3+i)^2)^3 = 1*(3+i)^6 = (3+i)^6$.
So if the three solutions to $w^3=1$ are $1, b,c$ then the three solutions to $z^3 = (3+i)^6$ are $8 + 6i, b(8+6i), c(8+6i)$.
So we nee to solve $w^3 = 1$.
Let $w = m+ni$ (where $m,n$ are real) then $(m+ni)^3 = m^3 + 3m^2ni + 3mn^2i^2 + n^3i^3=$
$m^3 + 3m^2ni -3mn^2 - n^3i = $
$(m^3 - 3mn^2) + (3m^2n-n^3)i = $
$1 + 0i$
So $m^3 -3mn^2 =1$ and $3m^2n -n^3=0$.
$3m^2n -n^3= n(3m^2-n^2) = n(msqrt 3 -n)(msqrt 3 + n) = 0$ so
either:
1) $n = 0$ and $m^3 = 1$ so $m = 1$ and $w = 1 + 0i = 1$.
2) $msqrt 3 - n = 0$ and $msqrt 3 = n$ so $m^3 -3mn^2 = 1$ and so
$m^3 - 3m(msqrt 3)^2 = m^3 - 3m(3m^2) =m^3 -9m^3 = -8m^3 =1$ so $m = -frac 12$ and $w = -frac 12 -frac {sqrt 3}2 i$.
3) $msqrt 3+ n = 0$ so $n = -msqrt 3$ and so
$m^3 - 3m(-msqrt 3)^2=.... = 1$ so $m = -frac 12$ and $w = -frac 12 + frac {sqrt 3}2i$.
So the three solutions to $z^3 =(3+i)^6$ are:
1)$8+6i$
2)$(-frac 12 -frac {sqrt 3}2 i)(8 + 6i) = -4 -4sqrt 3i - 3i -3sqrt 3 i^2 = (-4+3sqrt 3) -(4sqrt 3-3)i$.
and 3) $(-frac 12 +frac {sqrt 3}2 i)(8 + 6i) = -4 +4sqrt 3i - 3i +3sqrt 3 i^2 = (-4-3sqrt 3) +(4sqrt 3-3)i$
....
However once you learn polar notation that $z = m + ni = r(cos theta + isin theta) := re^{itheta}$ where $r = sqrt{m^2 + n^2}$ and $theta$ is the solutions to $cos theta = frac mr$ and $sin theta = frac nr$, once you learn that these problems become MUCH easier.
answered Jan 19 at 23:58
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
Whenever you have $z^3 = k$ there will be $3$ solutions.
If $z^3 = (3+i)^6 = ((3+i)^2)^3$ then one solution is obviously $z = (3+i)^2 = 9 + 6i + i^2 = 9+6i -1 = 8+6i$.
But that is only one solution.
However $w^3 = 1$ has three solutions.
If we let $w$ be one of the solutions and we let $z = w(8+6i) = w(3+i)^2$ then $z^3 = w^3((3+i)^2)^3 = 1*(3+i)^6 = (3+i)^6$.
So if the three solutions to $w^3=1$ are $1, b,c$ then the three solutions to $z^3 = (3+i)^6$ are $8 + 6i, b(8+6i), c(8+6i)$.
So we nee to solve $w^3 = 1$.
Let $w = m+ni$ (where $m,n$ are real) then $(m+ni)^3 = m^3 + 3m^2ni + 3mn^2i^2 + n^3i^3=$
$m^3 + 3m^2ni -3mn^2 - n^3i = $
$(m^3 - 3mn^2) + (3m^2n-n^3)i = $
$1 + 0i$
So $m^3 -3mn^2 =1$ and $3m^2n -n^3=0$.
$3m^2n -n^3= n(3m^2-n^2) = n(msqrt 3 -n)(msqrt 3 + n) = 0$ so
either:
1) $n = 0$ and $m^3 = 1$ so $m = 1$ and $w = 1 + 0i = 1$.
2) $msqrt 3 - n = 0$ and $msqrt 3 = n$ so $m^3 -3mn^2 = 1$ and so
$m^3 - 3m(msqrt 3)^2 = m^3 - 3m(3m^2) =m^3 -9m^3 = -8m^3 =1$ so $m = -frac 12$ and $w = -frac 12 -frac {sqrt 3}2 i$.
3) $msqrt 3+ n = 0$ so $n = -msqrt 3$ and so
$m^3 - 3m(-msqrt 3)^2=.... = 1$ so $m = -frac 12$ and $w = -frac 12 + frac {sqrt 3}2i$.
So the three solutions to $z^3 =(3+i)^6$ are:
1)$8+6i$
2)$(-frac 12 -frac {sqrt 3}2 i)(8 + 6i) = -4 -4sqrt 3i - 3i -3sqrt 3 i^2 = (-4+3sqrt 3) -(4sqrt 3-3)i$.
and 3) $(-frac 12 +frac {sqrt 3}2 i)(8 + 6i) = -4 +4sqrt 3i - 3i +3sqrt 3 i^2 = (-4-3sqrt 3) +(4sqrt 3-3)i$
....
However once you learn polar notation that $z = m + ni = r(cos theta + isin theta) := re^{itheta}$ where $r = sqrt{m^2 + n^2}$ and $theta$ is the solutions to $cos theta = frac mr$ and $sin theta = frac nr$, once you learn that these problems become MUCH easier.
answered Jan 19 at 23:58
fleabloodfleablood
71.7k22686
$endgroup$
Whenever you have $z^3 = k$ there will be $3$ solutions.
If $z^3 = (3+i)^6 = ((3+i)^2)^3$ then one solution is obviously $z = (3+i)^2 = 9 + 6i + i^2 = 9+6i -1 = 8+6i$.
But that is only one solution.
However $w^3 = 1$ has three solutions.
If we let $w$ be one of the solutions and we let $z = w(8+6i) = w(3+i)^2$ then $z^3 = w^3((3+i)^2)^3 = 1*(3+i)^6 = (3+i)^6$.
So if the three solutions to $w^3=1$ are $1, b,c$ then the three solutions to $z^3 = (3+i)^6$ are $8 + 6i, b(8+6i), c(8+6i)$.
So we nee to solve $w^3 = 1$.
Let $w = m+ni$ (where $m,n$ are real) then $(m+ni)^3 = m^3 + 3m^2ni + 3mn^2i^2 + n^3i^3=$
$m^3 + 3m^2ni -3mn^2 - n^3i = $
$(m^3 - 3mn^2) + (3m^2n-n^3)i = $
$1 + 0i$
So $m^3 -3mn^2 =1$ and $3m^2n -n^3=0$.
$3m^2n -n^3= n(3m^2-n^2) = n(msqrt 3 -n)(msqrt 3 + n) = 0$ so
either:
1) $n = 0$ and $m^3 = 1$ so $m = 1$ and $w = 1 + 0i = 1$.
2) $msqrt 3 - n = 0$ and $msqrt 3 = n$ so $m^3 -3mn^2 = 1$ and so
$m^3 - 3m(msqrt 3)^2 = m^3 - 3m(3m^2) =m^3 -9m^3 = -8m^3 =1$ so $m = -frac 12$ and $w = -frac 12 -frac {sqrt 3}2 i$.
3) $msqrt 3+ n = 0$ so $n = -msqrt 3$ and so
$m^3 - 3m(-msqrt 3)^2=.... = 1$ so $m = -frac 12$ and $w = -frac 12 + frac {sqrt 3}2i$.
So the three solutions to $z^3 =(3+i)^6$ are:
1)$8+6i$
2)$(-frac 12 -frac {sqrt 3}2 i)(8 + 6i) = -4 -4sqrt 3i - 3i -3sqrt 3 i^2 = (-4+3sqrt 3) -(4sqrt 3-3)i$.
and 3) $(-frac 12 +frac {sqrt 3}2 i)(8 + 6i) = -4 +4sqrt 3i - 3i +3sqrt 3 i^2 = (-4-3sqrt 3) +(4sqrt 3-3)i$
....
However once you learn polar notation that $z = m + ni = r(cos theta + isin theta) := re^{itheta}$ where $r = sqrt{m^2 + n^2}$ and $theta$ is the solutions to $cos theta = frac mr$ and $sin theta = frac nr$, once you learn that these problems become MUCH easier.
answered Jan 19 at 23:58
fleabloodfleablood
71.7k22686
answered Jan 19 at 23:58
fleabloodfleablood
71.7k22686
answered Jan 19 at 23:58
fleabloodfleablood
71.7k22686
answered Jan 19 at 23:58
fleabloodfleablood
71.7k22686
71.7k22686
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$begingroup$
Note that $z^3 = left(3 + iright)^6$, by moving the right side value to the left, basically forms a cubic equation in $z$. Thus, there's always $3$ roots, although some may be repeated. Have you accounted for this?
$endgroup$
– John Omielan
Jan 19 at 23:22
$begingroup$
As already mentioned there are three different candidates for $z$. One of them is found correctly.
$endgroup$
– user
Jan 19 at 23:26