Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$
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Problem
Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$
Attempt to solve
$$= sqrt2e^i 7pi/12-e^i pi / 3 = e^ifrac56pi$$
I can't seem to find how this expression is $e^ifrac56pi$ ?
I would like the final result be in polar form $re^itheta$
complex-numbers
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up vote
0
down vote
favorite
Problem
Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$
Attempt to solve
$$= sqrt2e^i 7pi/12-e^i pi / 3 = e^ifrac56pi$$
I can't seem to find how this expression is $e^ifrac56pi$ ?
I would like the final result be in polar form $re^itheta$
complex-numbers
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Problem
Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$
Attempt to solve
$$= sqrt2e^i 7pi/12-e^i pi / 3 = e^ifrac56pi$$
I can't seem to find how this expression is $e^ifrac56pi$ ?
I would like the final result be in polar form $re^itheta$
complex-numbers
Problem
Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$
Attempt to solve
$$= sqrt2e^i 7pi/12-e^i pi / 3 = e^ifrac56pi$$
I can't seem to find how this expression is $e^ifrac56pi$ ?
I would like the final result be in polar form $re^itheta$
complex-numbers
complex-numbers
asked Sep 2 at 9:55
Tuki
637316
637316
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2 Answers
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oldest
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up vote
3
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accepted
It might be slightly simpler to notice that
$$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
so
$$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$
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up vote
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One may write trigonometric form
beginalign
e^ifracpi3(sqrt2e^ifracpi4 -1)
&= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
&= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
&= dfrac-sqrt32+idfrac12 \
&= e^frac56pi i
endalign
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It might be slightly simpler to notice that
$$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
so
$$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$
add a comment |Â
up vote
3
down vote
accepted
It might be slightly simpler to notice that
$$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
so
$$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It might be slightly simpler to notice that
$$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
so
$$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$
It might be slightly simpler to notice that
$$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
so
$$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$
answered Sep 2 at 9:59
Sobi
2,843417
2,843417
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add a comment |Â
up vote
1
down vote
One may write trigonometric form
beginalign
e^ifracpi3(sqrt2e^ifracpi4 -1)
&= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
&= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
&= dfrac-sqrt32+idfrac12 \
&= e^frac56pi i
endalign
add a comment |Â
up vote
1
down vote
One may write trigonometric form
beginalign
e^ifracpi3(sqrt2e^ifracpi4 -1)
&= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
&= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
&= dfrac-sqrt32+idfrac12 \
&= e^frac56pi i
endalign
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One may write trigonometric form
beginalign
e^ifracpi3(sqrt2e^ifracpi4 -1)
&= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
&= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
&= dfrac-sqrt32+idfrac12 \
&= e^frac56pi i
endalign
One may write trigonometric form
beginalign
e^ifracpi3(sqrt2e^ifracpi4 -1)
&= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
&= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
&= dfrac-sqrt32+idfrac12 \
&= e^frac56pi i
endalign
answered Sep 2 at 10:15
Nosrati
22.1k61747
22.1k61747
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add a comment |Â
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