Mixing
Сomplex number equation: $z^3+frac{(sqrt2+sqrt2i)^7}{i^{11}(-6+2sqrt3i)^{13}}=0$
$begingroup$
please tell me how I can solve the following equation.
$$z^3+frac{(sqrt2+sqrt2i)^7}{i^{11}(-6+2sqrt3i)^{13}}=0$$
What formula should I use? If possible, tell me how to solve this equation or write where I can find a formula for solving such an equation. I searched for it on the Internet, but could not find anything useful.
complex-numbers
edited Jan 20 at 13:30
greedoid
45.8k1159116
asked Jan 20 at 13:23
Roman AtrashchenokRoman Atrashchenok
31
$endgroup$
add a comment |
$begingroup$
please tell me how I can solve the following equation.
$$z^3+frac{(sqrt2+sqrt2i)^7}{i^{11}(-6+2sqrt3i)^{13}}=0$$
What formula should I use? If possible, tell me how to solve this equation or write where I can find a formula for solving such an equation. I searched for it on the Internet, but could not find anything useful.
complex-numbers
edited Jan 20 at 13:30
greedoid
45.8k1159116
asked Jan 20 at 13:23
Roman AtrashchenokRoman Atrashchenok
31
$endgroup$
$begingroup$
I think you think that this is more difficult than it really is. Can you solve an equation like $z^3+2i=0$? Because this one isn't much different from that. And if you can't solve my simpler equation, then start there instead of with yours.
$endgroup$
– Arthur
Jan 20 at 13:25
$begingroup$
Write it in a polar form.
$endgroup$
– greedoid
Jan 20 at 13:31
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin Sleziak
Jan 20 at 13:32
add a comment |
$begingroup$
please tell me how I can solve the following equation.
$$z^3+frac{(sqrt2+sqrt2i)^7}{i^{11}(-6+2sqrt3i)^{13}}=0$$
What formula should I use? If possible, tell me how to solve this equation or write where I can find a formula for solving such an equation. I searched for it on the Internet, but could not find anything useful.
complex-numbers
edited Jan 20 at 13:30
greedoid
45.8k1159116
asked Jan 20 at 13:23
Roman AtrashchenokRoman Atrashchenok
31
$endgroup$
please tell me how I can solve the following equation.
$$z^3+frac{(sqrt2+sqrt2i)^7}{i^{11}(-6+2sqrt3i)^{13}}=0$$
What formula should I use? If possible, tell me how to solve this equation or write where I can find a formula for solving such an equation. I searched for it on the Internet, but could not find anything useful.
complex-numbers
complex-numbers
edited Jan 20 at 13:30
greedoid
45.8k1159116
asked Jan 20 at 13:23
Roman AtrashchenokRoman Atrashchenok
31
edited Jan 20 at 13:30
greedoid
45.8k1159116
asked Jan 20 at 13:23
Roman AtrashchenokRoman Atrashchenok
31
edited Jan 20 at 13:30
greedoid
45.8k1159116
edited Jan 20 at 13:30
greedoid
45.8k1159116
edited Jan 20 at 13:30
greedoid
45.8k1159116
45.8k1159116
asked Jan 20 at 13:23
Roman AtrashchenokRoman Atrashchenok
31
asked Jan 20 at 13:23
Roman AtrashchenokRoman Atrashchenok
31
asked Jan 20 at 13:23
Roman AtrashchenokRoman Atrashchenok
31
31
$begingroup$
I think you think that this is more difficult than it really is. Can you solve an equation like $z^3+2i=0$? Because this one isn't much different from that. And if you can't solve my simpler equation, then start there instead of with yours.
$endgroup$
– Arthur
Jan 20 at 13:25
$begingroup$
Write it in a polar form.
$endgroup$
– greedoid
Jan 20 at 13:31
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin Sleziak
Jan 20 at 13:32
add a comment |
$begingroup$
I think you think that this is more difficult than it really is. Can you solve an equation like $z^3+2i=0$? Because this one isn't much different from that. And if you can't solve my simpler equation, then start there instead of with yours.
$endgroup$
– Arthur
Jan 20 at 13:25
$begingroup$
Write it in a polar form.
$endgroup$
– greedoid
Jan 20 at 13:31
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin Sleziak
Jan 20 at 13:32
$begingroup$
I think you think that this is more difficult than it really is. Can you solve an equation like $z^3+2i=0$? Because this one isn't much different from that. And if you can't solve my simpler equation, then start there instead of with yours.
$endgroup$
– Arthur
Jan 20 at 13:25
$begingroup$
I think you think that this is more difficult than it really is. Can you solve an equation like $z^3+2i=0$? Because this one isn't much different from that. And if you can't solve my simpler equation, then start there instead of with yours.
$endgroup$
– Arthur
Jan 20 at 13:25
$begingroup$
Write it in a polar form.
$endgroup$
– greedoid
Jan 20 at 13:31
$begingroup$
Write it in a polar form.
$endgroup$
– greedoid
Jan 20 at 13:31
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin Sleziak
Jan 20 at 13:32
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin Sleziak
Jan 20 at 13:32
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$sqrt2+sqrt2i=2e^{frac{pi i}4}$.
And $-6+2sqrt3i=4sqrt3e^{frac{pi i}6}$.
And $i^{11}=-i$.
So we have $z^3-frac{2^7e^{frac{7pi i}4}}{icdot (4sqrt3)^{13}e^{frac{13pi i}6}}=0implies z^3+frac{i}{e^{frac{5pi i}{12}}2^{19}3^{frac{13}2}}=0implies z=-frac1{576cdot 2^{frac13}cdot 3^{frac16}}e^{frac{pi i}{36}},-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{2pi i}3}$ or $-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{4pi i}3}$.
answered Jan 20 at 14:39
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
Hint
You have to use the exponential form of complex numbers: set $Z=rmathrm e^{itheta}$ and observe that, for instance,
$$sqrt 2+sqrt 2i=sqrt 2(1+i)=sqrt 2cdot sqrt2mathrm e^{itfracpi 4}=2,mathrm e^{itfracpi 4}, ;text{&c.}$$
so that $;(sqrt 2+sqrt 2i)^7=2^7 mathrm e^{itfrac{7pi}4}$.
Can you proceed similarly for the denominator?
answered Jan 20 at 13:38
BernardBernard
122k740116
$endgroup$
$begingroup$
I will try, just this topic is new for me, I just started to study it today)
$endgroup$
– Roman Atrashchenok
Jan 20 at 14:06
add a comment |
$begingroup$
Noticing that $sqrt2+sqrt2i=2left(frac{sqrt2}2+ifrac{sqrt2}2right)$ and $-6+2sqrt3i=4sqrt3left(-frac{sqrt3}2+ifrac12right)$ should help you to express the given complex numbers in the polar form. You can then use de Moivre's formula to simplify the expression. From there you might be able to continue further.
answered Jan 20 at 13:41
Martin SleziakMartin Sleziak
44.8k10119272
$endgroup$
add a comment |
$begingroup$
Simplify at first, $$(sqrt{2}+sqrt{2}i)^7=64sqrt{2}(1-i)$$
to prove this write $$sqrt{2}^7(1+i)^7$$ Calculate $$(1+i)^3cdot (1+i)^3cdot (1+i)$$
answered Jan 20 at 13:28
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
I think the simplification requires a little more explanation.
$endgroup$
– Thomas Shelby
Jan 20 at 15:34
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$sqrt2+sqrt2i=2e^{frac{pi i}4}$.
And $-6+2sqrt3i=4sqrt3e^{frac{pi i}6}$.
And $i^{11}=-i$.
So we have $z^3-frac{2^7e^{frac{7pi i}4}}{icdot (4sqrt3)^{13}e^{frac{13pi i}6}}=0implies z^3+frac{i}{e^{frac{5pi i}{12}}2^{19}3^{frac{13}2}}=0implies z=-frac1{576cdot 2^{frac13}cdot 3^{frac16}}e^{frac{pi i}{36}},-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{2pi i}3}$ or $-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{4pi i}3}$.
answered Jan 20 at 14:39
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
$sqrt2+sqrt2i=2e^{frac{pi i}4}$.
And $-6+2sqrt3i=4sqrt3e^{frac{pi i}6}$.
And $i^{11}=-i$.
So we have $z^3-frac{2^7e^{frac{7pi i}4}}{icdot (4sqrt3)^{13}e^{frac{13pi i}6}}=0implies z^3+frac{i}{e^{frac{5pi i}{12}}2^{19}3^{frac{13}2}}=0implies z=-frac1{576cdot 2^{frac13}cdot 3^{frac16}}e^{frac{pi i}{36}},-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{2pi i}3}$ or $-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{4pi i}3}$.
answered Jan 20 at 14:39
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
$sqrt2+sqrt2i=2e^{frac{pi i}4}$.
And $-6+2sqrt3i=4sqrt3e^{frac{pi i}6}$.
And $i^{11}=-i$.
So we have $z^3-frac{2^7e^{frac{7pi i}4}}{icdot (4sqrt3)^{13}e^{frac{13pi i}6}}=0implies z^3+frac{i}{e^{frac{5pi i}{12}}2^{19}3^{frac{13}2}}=0implies z=-frac1{576cdot 2^{frac13}cdot 3^{frac16}}e^{frac{pi i}{36}},-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{2pi i}3}$ or $-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{4pi i}3}$.
answered Jan 20 at 14:39
Chris CusterChris Custer
14.1k3827
$endgroup$
$sqrt2+sqrt2i=2e^{frac{pi i}4}$.
And $-6+2sqrt3i=4sqrt3e^{frac{pi i}6}$.
And $i^{11}=-i$.
So we have $z^3-frac{2^7e^{frac{7pi i}4}}{icdot (4sqrt3)^{13}e^{frac{13pi i}6}}=0implies z^3+frac{i}{e^{frac{5pi i}{12}}2^{19}3^{frac{13}2}}=0implies z=-frac1{576cdot 2^{frac13}cdot 3^{frac16}}e^{frac{pi i}{36}},-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{2pi i}3}$ or $-frac{e^{frac{pi i}{36}}}{{576cdot2^{frac13}cdot 3^frac16}}cdot e^{frac{4pi i}3}$.
answered Jan 20 at 14:39
Chris CusterChris Custer
14.1k3827
edited Jan 23 at 2:32
answered Jan 20 at 14:39
Chris CusterChris Custer
14.1k3827
answered Jan 20 at 14:39
Chris CusterChris Custer
14.1k3827
answered Jan 20 at 14:39
Chris CusterChris Custer
14.1k3827
14.1k3827
add a comment |
add a comment |
$begingroup$
Hint
You have to use the exponential form of complex numbers: set $Z=rmathrm e^{itheta}$ and observe that, for instance,
$$sqrt 2+sqrt 2i=sqrt 2(1+i)=sqrt 2cdot sqrt2mathrm e^{itfracpi 4}=2,mathrm e^{itfracpi 4}, ;text{&c.}$$
so that $;(sqrt 2+sqrt 2i)^7=2^7 mathrm e^{itfrac{7pi}4}$.
Can you proceed similarly for the denominator?
answered Jan 20 at 13:38
BernardBernard
122k740116
$endgroup$
$begingroup$
I will try, just this topic is new for me, I just started to study it today)
$endgroup$
– Roman Atrashchenok
Jan 20 at 14:06
add a comment |
$begingroup$
Hint
You have to use the exponential form of complex numbers: set $Z=rmathrm e^{itheta}$ and observe that, for instance,
$$sqrt 2+sqrt 2i=sqrt 2(1+i)=sqrt 2cdot sqrt2mathrm e^{itfracpi 4}=2,mathrm e^{itfracpi 4}, ;text{&c.}$$
so that $;(sqrt 2+sqrt 2i)^7=2^7 mathrm e^{itfrac{7pi}4}$.
Can you proceed similarly for the denominator?
answered Jan 20 at 13:38
BernardBernard
122k740116
$endgroup$
$begingroup$
I will try, just this topic is new for me, I just started to study it today)
$endgroup$
– Roman Atrashchenok
Jan 20 at 14:06
add a comment |
$begingroup$
Hint
You have to use the exponential form of complex numbers: set $Z=rmathrm e^{itheta}$ and observe that, for instance,
$$sqrt 2+sqrt 2i=sqrt 2(1+i)=sqrt 2cdot sqrt2mathrm e^{itfracpi 4}=2,mathrm e^{itfracpi 4}, ;text{&c.}$$
so that $;(sqrt 2+sqrt 2i)^7=2^7 mathrm e^{itfrac{7pi}4}$.
Can you proceed similarly for the denominator?
answered Jan 20 at 13:38
BernardBernard
122k740116
$endgroup$
Hint
You have to use the exponential form of complex numbers: set $Z=rmathrm e^{itheta}$ and observe that, for instance,
$$sqrt 2+sqrt 2i=sqrt 2(1+i)=sqrt 2cdot sqrt2mathrm e^{itfracpi 4}=2,mathrm e^{itfracpi 4}, ;text{&c.}$$
so that $;(sqrt 2+sqrt 2i)^7=2^7 mathrm e^{itfrac{7pi}4}$.
Can you proceed similarly for the denominator?
answered Jan 20 at 13:38
BernardBernard
122k740116
answered Jan 20 at 13:38
BernardBernard
122k740116
answered Jan 20 at 13:38
BernardBernard
122k740116
answered Jan 20 at 13:38
BernardBernard
122k740116
122k740116
$begingroup$
I will try, just this topic is new for me, I just started to study it today)
$endgroup$
– Roman Atrashchenok
Jan 20 at 14:06
add a comment |
$begingroup$
I will try, just this topic is new for me, I just started to study it today)
$endgroup$
– Roman Atrashchenok
Jan 20 at 14:06
$begingroup$
I will try, just this topic is new for me, I just started to study it today)
$endgroup$
– Roman Atrashchenok
Jan 20 at 14:06
$begingroup$
I will try, just this topic is new for me, I just started to study it today)
$endgroup$
– Roman Atrashchenok
Jan 20 at 14:06
add a comment |
$begingroup$
Noticing that $sqrt2+sqrt2i=2left(frac{sqrt2}2+ifrac{sqrt2}2right)$ and $-6+2sqrt3i=4sqrt3left(-frac{sqrt3}2+ifrac12right)$ should help you to express the given complex numbers in the polar form. You can then use de Moivre's formula to simplify the expression. From there you might be able to continue further.
answered Jan 20 at 13:41
Martin SleziakMartin Sleziak
44.8k10119272
$endgroup$
add a comment |
$begingroup$
Noticing that $sqrt2+sqrt2i=2left(frac{sqrt2}2+ifrac{sqrt2}2right)$ and $-6+2sqrt3i=4sqrt3left(-frac{sqrt3}2+ifrac12right)$ should help you to express the given complex numbers in the polar form. You can then use de Moivre's formula to simplify the expression. From there you might be able to continue further.
answered Jan 20 at 13:41
Martin SleziakMartin Sleziak
44.8k10119272
$endgroup$
add a comment |
$begingroup$
Noticing that $sqrt2+sqrt2i=2left(frac{sqrt2}2+ifrac{sqrt2}2right)$ and $-6+2sqrt3i=4sqrt3left(-frac{sqrt3}2+ifrac12right)$ should help you to express the given complex numbers in the polar form. You can then use de Moivre's formula to simplify the expression. From there you might be able to continue further.
answered Jan 20 at 13:41
Martin SleziakMartin Sleziak
44.8k10119272
$endgroup$
Noticing that $sqrt2+sqrt2i=2left(frac{sqrt2}2+ifrac{sqrt2}2right)$ and $-6+2sqrt3i=4sqrt3left(-frac{sqrt3}2+ifrac12right)$ should help you to express the given complex numbers in the polar form. You can then use de Moivre's formula to simplify the expression. From there you might be able to continue further.
answered Jan 20 at 13:41
Martin SleziakMartin Sleziak
44.8k10119272
edited Jan 20 at 15:25
answered Jan 20 at 13:41
Martin SleziakMartin Sleziak
44.8k10119272
answered Jan 20 at 13:41
Martin SleziakMartin Sleziak
44.8k10119272
answered Jan 20 at 13:41
Martin SleziakMartin Sleziak
44.8k10119272
44.8k10119272
add a comment |
add a comment |
$begingroup$
Simplify at first, $$(sqrt{2}+sqrt{2}i)^7=64sqrt{2}(1-i)$$
to prove this write $$sqrt{2}^7(1+i)^7$$ Calculate $$(1+i)^3cdot (1+i)^3cdot (1+i)$$
answered Jan 20 at 13:28
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
I think the simplification requires a little more explanation.
$endgroup$
– Thomas Shelby
Jan 20 at 15:34
add a comment |
$begingroup$
Simplify at first, $$(sqrt{2}+sqrt{2}i)^7=64sqrt{2}(1-i)$$
to prove this write $$sqrt{2}^7(1+i)^7$$ Calculate $$(1+i)^3cdot (1+i)^3cdot (1+i)$$
answered Jan 20 at 13:28
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
I think the simplification requires a little more explanation.
$endgroup$
– Thomas Shelby
Jan 20 at 15:34
add a comment |
$begingroup$
Simplify at first, $$(sqrt{2}+sqrt{2}i)^7=64sqrt{2}(1-i)$$
to prove this write $$sqrt{2}^7(1+i)^7$$ Calculate $$(1+i)^3cdot (1+i)^3cdot (1+i)$$
answered Jan 20 at 13:28
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
Simplify at first, $$(sqrt{2}+sqrt{2}i)^7=64sqrt{2}(1-i)$$
to prove this write $$sqrt{2}^7(1+i)^7$$ Calculate $$(1+i)^3cdot (1+i)^3cdot (1+i)$$
answered Jan 20 at 13:28
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
edited Jan 20 at 15:45
answered Jan 20 at 13:28
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Jan 20 at 13:28
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Jan 20 at 13:28
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
76.8k42866
$begingroup$
I think the simplification requires a little more explanation.
$endgroup$
– Thomas Shelby
Jan 20 at 15:34
add a comment |
$begingroup$
I think the simplification requires a little more explanation.
$endgroup$
– Thomas Shelby
Jan 20 at 15:34
$begingroup$
I think the simplification requires a little more explanation.
$endgroup$
– Thomas Shelby
Jan 20 at 15:34
$begingroup$
I think the simplification requires a little more explanation.
$endgroup$
– Thomas Shelby
Jan 20 at 15:34
add a comment |
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$begingroup$
I think you think that this is more difficult than it really is. Can you solve an equation like $z^3+2i=0$? Because this one isn't much different from that. And if you can't solve my simpler equation, then start there instead of with yours.
$endgroup$
– Arthur
Jan 20 at 13:25
$begingroup$
Write it in a polar form.
$endgroup$
– greedoid
Jan 20 at 13:31
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin Sleziak
Jan 20 at 13:32