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Calculating the given power of a complex number
The 2019 Stack Overflow Developer Survey Results Are InProblems with trigonometry getting the power of this complex expressionFind the matrix that corresponds to the composite transformation of a rotation followed by a translation.If $1+x^2=sqrt3x$ then $sum _n=1^24left(x^n+frac1x^nright)^2$ is equal toLimit of a complex number raised to a powerTrigonometric functions and complex numbersWhen changing the sign of the modulus of a complex number why do I have to change the sign of the argument too?Converting Complex numbers into Cartesian FormRelation between Circular functions and complex numbers.Linear goniometric equations $sin(x)=cos(x)$Absolute value of complex exponential divided by complex number
$begingroup$
$(-2sqrt3+2i)^-9$
I tried solving it like I solved other problems, I calculated $r$ which is $4$ then got $4(sinfracpi6+cosfracpi6)$, then using an equation,
I got $4^-9(sin90+cos90)=4^-9(1+0)=4^-9$
But the answer is $fraci2^18$.
I understand that $frac14^9$ is equal to $frac12^18$ but I don't understand how we got an $i$ in the answer.
Can anyone explain why?
linear-algebra complex-numbers
edited Mar 23 at 17:51
Rócherz
3,0263823
asked Mar 23 at 16:46
McAMcA
1
$endgroup$
add a comment |
$begingroup$
$(-2sqrt3+2i)^-9$
I tried solving it like I solved other problems, I calculated $r$ which is $4$ then got $4(sinfracpi6+cosfracpi6)$, then using an equation,
I got $4^-9(sin90+cos90)=4^-9(1+0)=4^-9$
But the answer is $fraci2^18$.
I understand that $frac14^9$ is equal to $frac12^18$ but I don't understand how we got an $i$ in the answer.
Can anyone explain why?
linear-algebra complex-numbers
edited Mar 23 at 17:51
Rócherz
3,0263823
asked Mar 23 at 16:46
McAMcA
1
$endgroup$
$begingroup$
Pleas see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Mar 23 at 16:48
add a comment |
$begingroup$
$(-2sqrt3+2i)^-9$
I tried solving it like I solved other problems, I calculated $r$ which is $4$ then got $4(sinfracpi6+cosfracpi6)$, then using an equation,
I got $4^-9(sin90+cos90)=4^-9(1+0)=4^-9$
But the answer is $fraci2^18$.
I understand that $frac14^9$ is equal to $frac12^18$ but I don't understand how we got an $i$ in the answer.
Can anyone explain why?
linear-algebra complex-numbers
edited Mar 23 at 17:51
Rócherz
3,0263823
asked Mar 23 at 16:46
McAMcA
1
$endgroup$
$(-2sqrt3+2i)^-9$
I tried solving it like I solved other problems, I calculated $r$ which is $4$ then got $4(sinfracpi6+cosfracpi6)$, then using an equation,
I got $4^-9(sin90+cos90)=4^-9(1+0)=4^-9$
But the answer is $fraci2^18$.
I understand that $frac14^9$ is equal to $frac12^18$ but I don't understand how we got an $i$ in the answer.
Can anyone explain why?
linear-algebra complex-numbers
linear-algebra complex-numbers
edited Mar 23 at 17:51
Rócherz
3,0263823
asked Mar 23 at 16:46
McAMcA
1
edited Mar 23 at 17:51
Rócherz
3,0263823
asked Mar 23 at 16:46
McAMcA
1
edited Mar 23 at 17:51
Rócherz
3,0263823
edited Mar 23 at 17:51
Rócherz
3,0263823
edited Mar 23 at 17:51
Rócherz
3,0263823
3,0263823
asked Mar 23 at 16:46
McAMcA
1
asked Mar 23 at 16:46
McAMcA
1
asked Mar 23 at 16:46
McAMcA
1
1
$begingroup$
Pleas see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Mar 23 at 16:48
add a comment |
$begingroup$
Pleas see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Mar 23 at 16:48
$begingroup$
Pleas see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Mar 23 at 16:48
$begingroup$
Pleas see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Mar 23 at 16:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I presume the question is to find
$$(2sqrt3+2i)^-9.$$
This equals
$$left(4left(cosfracpi6+isinfracpi6right)right)^-9
=2^-18left(cosleft(-frac3pi2right)
+isinleft(-frac3pi2right)right)=2^-18i$$
via de Moivre's theorem.
answered Mar 23 at 16:52
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Thanks for the solution but may I ask how you got the i in your answer? I solved some other problems too but they didn't have i in the answer.
$endgroup$
– McA
Mar 23 at 16:54
$begingroup$
de Moivre's theorem : $(cos t+isin t)^n=cos nt+isin nt$.
$endgroup$
– Lord Shark the Unknown
Mar 23 at 17:02
$begingroup$
Okay that makes sense I checked other problems and their sin was always zero. Thank you!
$endgroup$
– McA
Mar 23 at 17:03
add a comment |
Your Answer
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1 Answer
1
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1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
I presume the question is to find
$$(2sqrt3+2i)^-9.$$
This equals
$$left(4left(cosfracpi6+isinfracpi6right)right)^-9
=2^-18left(cosleft(-frac3pi2right)
+isinleft(-frac3pi2right)right)=2^-18i$$
via de Moivre's theorem.
answered Mar 23 at 16:52
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Thanks for the solution but may I ask how you got the i in your answer? I solved some other problems too but they didn't have i in the answer.
$endgroup$
– McA
Mar 23 at 16:54
$begingroup$
de Moivre's theorem : $(cos t+isin t)^n=cos nt+isin nt$.
$endgroup$
– Lord Shark the Unknown
Mar 23 at 17:02
$begingroup$
Okay that makes sense I checked other problems and their sin was always zero. Thank you!
$endgroup$
– McA
Mar 23 at 17:03
add a comment |
$begingroup$
I presume the question is to find
$$(2sqrt3+2i)^-9.$$
This equals
$$left(4left(cosfracpi6+isinfracpi6right)right)^-9
=2^-18left(cosleft(-frac3pi2right)
+isinleft(-frac3pi2right)right)=2^-18i$$
via de Moivre's theorem.
answered Mar 23 at 16:52
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Thanks for the solution but may I ask how you got the i in your answer? I solved some other problems too but they didn't have i in the answer.
$endgroup$
– McA
Mar 23 at 16:54
$begingroup$
de Moivre's theorem : $(cos t+isin t)^n=cos nt+isin nt$.
$endgroup$
– Lord Shark the Unknown
Mar 23 at 17:02
$begingroup$
Okay that makes sense I checked other problems and their sin was always zero. Thank you!
$endgroup$
– McA
Mar 23 at 17:03
add a comment |
$begingroup$
I presume the question is to find
$$(2sqrt3+2i)^-9.$$
This equals
$$left(4left(cosfracpi6+isinfracpi6right)right)^-9
=2^-18left(cosleft(-frac3pi2right)
+isinleft(-frac3pi2right)right)=2^-18i$$
via de Moivre's theorem.
answered Mar 23 at 16:52
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
I presume the question is to find
$$(2sqrt3+2i)^-9.$$
This equals
$$left(4left(cosfracpi6+isinfracpi6right)right)^-9
=2^-18left(cosleft(-frac3pi2right)
+isinleft(-frac3pi2right)right)=2^-18i$$
via de Moivre's theorem.
answered Mar 23 at 16:52
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 23 at 16:52
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 23 at 16:52
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 23 at 16:52
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
$begingroup$
Thanks for the solution but may I ask how you got the i in your answer? I solved some other problems too but they didn't have i in the answer.
$endgroup$
– McA
Mar 23 at 16:54
$begingroup$
de Moivre's theorem : $(cos t+isin t)^n=cos nt+isin nt$.
$endgroup$
– Lord Shark the Unknown
Mar 23 at 17:02
$begingroup$
Okay that makes sense I checked other problems and their sin was always zero. Thank you!
$endgroup$
– McA
Mar 23 at 17:03
add a comment |
$begingroup$
Thanks for the solution but may I ask how you got the i in your answer? I solved some other problems too but they didn't have i in the answer.
$endgroup$
– McA
Mar 23 at 16:54
$begingroup$
de Moivre's theorem : $(cos t+isin t)^n=cos nt+isin nt$.
$endgroup$
– Lord Shark the Unknown
Mar 23 at 17:02
$begingroup$
Okay that makes sense I checked other problems and their sin was always zero. Thank you!
$endgroup$
– McA
Mar 23 at 17:03
$begingroup$
Thanks for the solution but may I ask how you got the i in your answer? I solved some other problems too but they didn't have i in the answer.
$endgroup$
– McA
Mar 23 at 16:54
$begingroup$
Thanks for the solution but may I ask how you got the i in your answer? I solved some other problems too but they didn't have i in the answer.
$endgroup$
– McA
Mar 23 at 16:54
$begingroup$
de Moivre's theorem : $(cos t+isin t)^n=cos nt+isin nt$.
$endgroup$
– Lord Shark the Unknown
Mar 23 at 17:02
$begingroup$
de Moivre's theorem : $(cos t+isin t)^n=cos nt+isin nt$.
$endgroup$
– Lord Shark the Unknown
Mar 23 at 17:02
$begingroup$
Okay that makes sense I checked other problems and their sin was always zero. Thank you!
$endgroup$
– McA
Mar 23 at 17:03
$begingroup$
Okay that makes sense I checked other problems and their sin was always zero. Thank you!
$endgroup$
– McA
Mar 23 at 17:03
add a comment |
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$begingroup$
Pleas see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Mar 23 at 16:48