Complex Numbers in Polar Form raised to a power
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[1 + cos(ÃÂ/6) + isin(ÃÂ/6)]^6
For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(ÃÂ/6)), I'm not sure if the rule will apply.
How do I approach this?
complex-numbers
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up vote
3
down vote
favorite
[1 + cos(ÃÂ/6) + isin(ÃÂ/6)]^6
For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(ÃÂ/6)), I'm not sure if the rule will apply.
How do I approach this?
complex-numbers
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
[1 + cos(ÃÂ/6) + isin(ÃÂ/6)]^6
For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(ÃÂ/6)), I'm not sure if the rule will apply.
How do I approach this?
complex-numbers
[1 + cos(ÃÂ/6) + isin(ÃÂ/6)]^6
For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(ÃÂ/6)), I'm not sure if the rule will apply.
How do I approach this?
complex-numbers
complex-numbers
edited 1 hour ago
asked 1 hour ago
inspiration-less
213
213
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3 Answers
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Hint : remark that, for any $theta in mathbbR$, we have
$$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$
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Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.
Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.
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Hint:
To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
beginalign
(1&+u)^6+(1+bar u)^6= \
2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
endalign
You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.
Edit:
We can use the same method (expanding the binomial) to the new formulation of the question:
$$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
Note in addition to the above that:
$u+u^5=2operatornameIm u=i$,
$u^2+u^4=2operatornameIm u^2=isqrt 3$,
so $;(1+u)^6=1+(26+15sqrt 3)i.$
That's not the stated sum!
â Parcly Taxel
42 mins ago
@ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhileâ¦
â Bernard
36 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint : remark that, for any $theta in mathbbR$, we have
$$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$
add a comment |Â
up vote
3
down vote
Hint : remark that, for any $theta in mathbbR$, we have
$$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hint : remark that, for any $theta in mathbbR$, we have
$$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$
Hint : remark that, for any $theta in mathbbR$, we have
$$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$
answered 51 mins ago
S. Cho
15610
15610
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up vote
3
down vote
Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.
Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.
add a comment |Â
up vote
3
down vote
Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.
Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.
Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.
Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.
Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.
answered 50 mins ago
Parcly Taxel
34.3k136890
34.3k136890
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add a comment |Â
up vote
0
down vote
Hint:
To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
beginalign
(1&+u)^6+(1+bar u)^6= \
2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
endalign
You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.
Edit:
We can use the same method (expanding the binomial) to the new formulation of the question:
$$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
Note in addition to the above that:
$u+u^5=2operatornameIm u=i$,
$u^2+u^4=2operatornameIm u^2=isqrt 3$,
so $;(1+u)^6=1+(26+15sqrt 3)i.$
That's not the stated sum!
â Parcly Taxel
42 mins ago
@ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhileâ¦
â Bernard
36 mins ago
add a comment |Â
up vote
0
down vote
Hint:
To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
beginalign
(1&+u)^6+(1+bar u)^6= \
2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
endalign
You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.
Edit:
We can use the same method (expanding the binomial) to the new formulation of the question:
$$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
Note in addition to the above that:
$u+u^5=2operatornameIm u=i$,
$u^2+u^4=2operatornameIm u^2=isqrt 3$,
so $;(1+u)^6=1+(26+15sqrt 3)i.$
That's not the stated sum!
â Parcly Taxel
42 mins ago
@ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhileâ¦
â Bernard
36 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
beginalign
(1&+u)^6+(1+bar u)^6= \
2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
endalign
You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.
Edit:
We can use the same method (expanding the binomial) to the new formulation of the question:
$$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
Note in addition to the above that:
$u+u^5=2operatornameIm u=i$,
$u^2+u^4=2operatornameIm u^2=isqrt 3$,
so $;(1+u)^6=1+(26+15sqrt 3)i.$
Hint:
To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
beginalign
(1&+u)^6+(1+bar u)^6= \
2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
endalign
You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.
Edit:
We can use the same method (expanding the binomial) to the new formulation of the question:
$$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
Note in addition to the above that:
$u+u^5=2operatornameIm u=i$,
$u^2+u^4=2operatornameIm u^2=isqrt 3$,
so $;(1+u)^6=1+(26+15sqrt 3)i.$
edited 18 mins ago
answered 48 mins ago
Bernard
113k636104
113k636104
That's not the stated sum!
â Parcly Taxel
42 mins ago
@ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhileâ¦
â Bernard
36 mins ago
add a comment |Â
That's not the stated sum!
â Parcly Taxel
42 mins ago
@ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhileâ¦
â Bernard
36 mins ago
That's not the stated sum!
â Parcly Taxel
42 mins ago
That's not the stated sum!
â Parcly Taxel
42 mins ago
@ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhileâ¦
â Bernard
36 mins ago
@ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhileâ¦
â Bernard
36 mins ago
add a comment |Â
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