Mixing
Using de Moivre to solve $z^3=-1$, one solution is $cos pi+i sin pi$. What am I doing wrong?
$begingroup$
I have a problem in my calculations using de Moivre for complex numbers: I have been staring at this for an hour, without success. Can anybody please tell me what I am doing wrong?
I have to calculate $z^3=-1$
So, besides the obvious solution $z=-1$:
$$z^3=r^3(cos3theta+i sin3theta)=1(cospi + i sin pi)$$
so $r = 1$ and $3theta=pi+2kpi$, so
$$θ = frac{pi}{3}+frac{2}{3}kpi$$
The first solution that comes out of this is correct, namely $cos frac{pi}{3}+i sin frac{pi}{3}$. However, the second solution that comes out is patently wrong: $cos pi+i sin pi$. The third one would be right again: $cos frac{5pi}{3}+i sin frac{5pi}{3}$, which would be our final one. I do not understand why the second one came out wrong and what I should change in the procedure so that I do not make this mistake again.
The error is obviously located somewhere in the $θ = frac{π}{3}+frac{2}{3}kπ$, but this is how the textbook instructs me to go through the procedure.
Any thoughts would be really appreciated...
trigonometry complex-numbers
edited Feb 2 at 22:42
Blue
49.7k870158
asked Feb 2 at 16:49
daltadalta
1578
$endgroup$
add a comment |
$begingroup$
I have a problem in my calculations using de Moivre for complex numbers: I have been staring at this for an hour, without success. Can anybody please tell me what I am doing wrong?
I have to calculate $z^3=-1$
So, besides the obvious solution $z=-1$:
$$z^3=r^3(cos3theta+i sin3theta)=1(cospi + i sin pi)$$
so $r = 1$ and $3theta=pi+2kpi$, so
$$θ = frac{pi}{3}+frac{2}{3}kpi$$
The first solution that comes out of this is correct, namely $cos frac{pi}{3}+i sin frac{pi}{3}$. However, the second solution that comes out is patently wrong: $cos pi+i sin pi$. The third one would be right again: $cos frac{5pi}{3}+i sin frac{5pi}{3}$, which would be our final one. I do not understand why the second one came out wrong and what I should change in the procedure so that I do not make this mistake again.
The error is obviously located somewhere in the $θ = frac{π}{3}+frac{2}{3}kπ$, but this is how the textbook instructs me to go through the procedure.
Any thoughts would be really appreciated...
trigonometry complex-numbers
edited Feb 2 at 22:42
Blue
49.7k870158
asked Feb 2 at 16:49
daltadalta
1578
$endgroup$
1
$begingroup$
Where is the problem? Your second solution is -1.
$endgroup$
– Exp ikx
Feb 2 at 16:54
$begingroup$
Sorry, I am totally new to this: now I see it…!
$endgroup$
– dalta
Feb 2 at 16:54
add a comment |
$begingroup$
I have a problem in my calculations using de Moivre for complex numbers: I have been staring at this for an hour, without success. Can anybody please tell me what I am doing wrong?
I have to calculate $z^3=-1$
So, besides the obvious solution $z=-1$:
$$z^3=r^3(cos3theta+i sin3theta)=1(cospi + i sin pi)$$
so $r = 1$ and $3theta=pi+2kpi$, so
$$θ = frac{pi}{3}+frac{2}{3}kpi$$
The first solution that comes out of this is correct, namely $cos frac{pi}{3}+i sin frac{pi}{3}$. However, the second solution that comes out is patently wrong: $cos pi+i sin pi$. The third one would be right again: $cos frac{5pi}{3}+i sin frac{5pi}{3}$, which would be our final one. I do not understand why the second one came out wrong and what I should change in the procedure so that I do not make this mistake again.
The error is obviously located somewhere in the $θ = frac{π}{3}+frac{2}{3}kπ$, but this is how the textbook instructs me to go through the procedure.
Any thoughts would be really appreciated...
trigonometry complex-numbers
edited Feb 2 at 22:42
Blue
49.7k870158
asked Feb 2 at 16:49
daltadalta
1578
$endgroup$
I have a problem in my calculations using de Moivre for complex numbers: I have been staring at this for an hour, without success. Can anybody please tell me what I am doing wrong?
I have to calculate $z^3=-1$
So, besides the obvious solution $z=-1$:
$$z^3=r^3(cos3theta+i sin3theta)=1(cospi + i sin pi)$$
so $r = 1$ and $3theta=pi+2kpi$, so
$$θ = frac{pi}{3}+frac{2}{3}kpi$$
The first solution that comes out of this is correct, namely $cos frac{pi}{3}+i sin frac{pi}{3}$. However, the second solution that comes out is patently wrong: $cos pi+i sin pi$. The third one would be right again: $cos frac{5pi}{3}+i sin frac{5pi}{3}$, which would be our final one. I do not understand why the second one came out wrong and what I should change in the procedure so that I do not make this mistake again.
The error is obviously located somewhere in the $θ = frac{π}{3}+frac{2}{3}kπ$, but this is how the textbook instructs me to go through the procedure.
Any thoughts would be really appreciated...
trigonometry complex-numbers
trigonometry complex-numbers
edited Feb 2 at 22:42
Blue
49.7k870158
asked Feb 2 at 16:49
daltadalta
1578
edited Feb 2 at 22:42
Blue
49.7k870158
asked Feb 2 at 16:49
daltadalta
1578
edited Feb 2 at 22:42
Blue
49.7k870158
edited Feb 2 at 22:42
Blue
49.7k870158
edited Feb 2 at 22:42
Blue
49.7k870158
49.7k870158
asked Feb 2 at 16:49
daltadalta
1578
asked Feb 2 at 16:49
daltadalta
1578
asked Feb 2 at 16:49
daltadalta
1578
1578
1
$begingroup$
Where is the problem? Your second solution is -1.
$endgroup$
– Exp ikx
Feb 2 at 16:54
$begingroup$
Sorry, I am totally new to this: now I see it…!
$endgroup$
– dalta
Feb 2 at 16:54
add a comment |
1
$begingroup$
Where is the problem? Your second solution is -1.
$endgroup$
– Exp ikx
Feb 2 at 16:54
$begingroup$
Sorry, I am totally new to this: now I see it…!
$endgroup$
– dalta
Feb 2 at 16:54
1
1
$begingroup$
Where is the problem? Your second solution is -1.
$endgroup$
– Exp ikx
Feb 2 at 16:54
$begingroup$
Where is the problem? Your second solution is -1.
$endgroup$
– Exp ikx
Feb 2 at 16:54
$begingroup$
Sorry, I am totally new to this: now I see it…!
$endgroup$
– dalta
Feb 2 at 16:54
$begingroup$
Sorry, I am totally new to this: now I see it…!
$endgroup$
– dalta
Feb 2 at 16:54
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Since $cos(pi)+isin(pi)=-1$ and since you wrote that $-1$ is a solution, I fail to see what's the problem.
answered Feb 2 at 16:51
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Oh wow, I get it now… Sorry: I am totally new to this!
$endgroup$
– dalta
Feb 2 at 16:54
add a comment |
$begingroup$
The second solution ($cos pi + i sin pi$) is correct (part of the root for $z^3 =-1$) as $cos pi$ is $-1$ and $sin pi$ is $0$, hence you have $-1+0i$ = $-1.$ From a unit circle perspective, $cos theta$ is decreasing and negative and $sin theta$ is decreasing but positive in the second quadrant.
answered Feb 2 at 23:24
bjcolby15bjcolby15
1,51511016
$endgroup$
add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
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$begingroup$
Since $cos(pi)+isin(pi)=-1$ and since you wrote that $-1$ is a solution, I fail to see what's the problem.
answered Feb 2 at 16:51
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Oh wow, I get it now… Sorry: I am totally new to this!
$endgroup$
– dalta
Feb 2 at 16:54
add a comment |
$begingroup$
Since $cos(pi)+isin(pi)=-1$ and since you wrote that $-1$ is a solution, I fail to see what's the problem.
answered Feb 2 at 16:51
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Oh wow, I get it now… Sorry: I am totally new to this!
$endgroup$
– dalta
Feb 2 at 16:54
add a comment |
$begingroup$
Since $cos(pi)+isin(pi)=-1$ and since you wrote that $-1$ is a solution, I fail to see what's the problem.
answered Feb 2 at 16:51
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
Since $cos(pi)+isin(pi)=-1$ and since you wrote that $-1$ is a solution, I fail to see what's the problem.
answered Feb 2 at 16:51
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 16:51
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 16:51
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 16:51
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
$begingroup$
Oh wow, I get it now… Sorry: I am totally new to this!
$endgroup$
– dalta
Feb 2 at 16:54
add a comment |
$begingroup$
Oh wow, I get it now… Sorry: I am totally new to this!
$endgroup$
– dalta
Feb 2 at 16:54
$begingroup$
Oh wow, I get it now… Sorry: I am totally new to this!
$endgroup$
– dalta
Feb 2 at 16:54
$begingroup$
Oh wow, I get it now… Sorry: I am totally new to this!
$endgroup$
– dalta
Feb 2 at 16:54
add a comment |
$begingroup$
The second solution ($cos pi + i sin pi$) is correct (part of the root for $z^3 =-1$) as $cos pi$ is $-1$ and $sin pi$ is $0$, hence you have $-1+0i$ = $-1.$ From a unit circle perspective, $cos theta$ is decreasing and negative and $sin theta$ is decreasing but positive in the second quadrant.
answered Feb 2 at 23:24
bjcolby15bjcolby15
1,51511016
$endgroup$
add a comment |
$begingroup$
The second solution ($cos pi + i sin pi$) is correct (part of the root for $z^3 =-1$) as $cos pi$ is $-1$ and $sin pi$ is $0$, hence you have $-1+0i$ = $-1.$ From a unit circle perspective, $cos theta$ is decreasing and negative and $sin theta$ is decreasing but positive in the second quadrant.
answered Feb 2 at 23:24
bjcolby15bjcolby15
1,51511016
$endgroup$
add a comment |
$begingroup$
The second solution ($cos pi + i sin pi$) is correct (part of the root for $z^3 =-1$) as $cos pi$ is $-1$ and $sin pi$ is $0$, hence you have $-1+0i$ = $-1.$ From a unit circle perspective, $cos theta$ is decreasing and negative and $sin theta$ is decreasing but positive in the second quadrant.
answered Feb 2 at 23:24
bjcolby15bjcolby15
1,51511016
$endgroup$
The second solution ($cos pi + i sin pi$) is correct (part of the root for $z^3 =-1$) as $cos pi$ is $-1$ and $sin pi$ is $0$, hence you have $-1+0i$ = $-1.$ From a unit circle perspective, $cos theta$ is decreasing and negative and $sin theta$ is decreasing but positive in the second quadrant.
answered Feb 2 at 23:24
bjcolby15bjcolby15
1,51511016
answered Feb 2 at 23:24
bjcolby15bjcolby15
1,51511016
answered Feb 2 at 23:24
bjcolby15bjcolby15
1,51511016
answered Feb 2 at 23:24
bjcolby15bjcolby15
1,51511016
1,51511016
add a comment |
add a comment |
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1
$begingroup$
Where is the problem? Your second solution is -1.
$endgroup$
– Exp ikx
Feb 2 at 16:54
$begingroup$
Sorry, I am totally new to this: now I see it…!
$endgroup$
– dalta
Feb 2 at 16:54