Mixing
Finding a complex number
$begingroup$
So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.
Here is the illustrative picture that I have:
I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$
Are my calculations right?
complex-numbers
asked Jan 7 at 21:19
Student123Student123
536
$endgroup$
add a comment |
$begingroup$
So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.
Here is the illustrative picture that I have:
I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$
Are my calculations right?
complex-numbers
asked Jan 7 at 21:19
Student123Student123
536
$endgroup$
$begingroup$
Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
$endgroup$
– EdOverflow
Jan 7 at 21:45
$begingroup$
Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
$endgroup$
– Emilio Novati
Jan 7 at 21:51
$begingroup$
no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
$endgroup$
– Student123
Jan 7 at 21:59
add a comment |
$begingroup$
So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.
Here is the illustrative picture that I have:
I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$
Are my calculations right?
complex-numbers
asked Jan 7 at 21:19
Student123Student123
536
$endgroup$
So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.
Here is the illustrative picture that I have:
I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$
Are my calculations right?
complex-numbers
complex-numbers
asked Jan 7 at 21:19
Student123Student123
536
asked Jan 7 at 21:19
Student123Student123
536
edited Jan 7 at 22:27
Student123
asked Jan 7 at 21:19
Student123Student123
536
asked Jan 7 at 21:19
Student123Student123
536
asked Jan 7 at 21:19
Student123Student123
536
536
$begingroup$
Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
$endgroup$
– EdOverflow
Jan 7 at 21:45
$begingroup$
Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
$endgroup$
– Emilio Novati
Jan 7 at 21:51
$begingroup$
no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
$endgroup$
– Student123
Jan 7 at 21:59
add a comment |
$begingroup$
Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
$endgroup$
– EdOverflow
Jan 7 at 21:45
$begingroup$
Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
$endgroup$
– Emilio Novati
Jan 7 at 21:51
$begingroup$
no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
$endgroup$
– Student123
Jan 7 at 21:59
$begingroup$
Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
$endgroup$
– EdOverflow
Jan 7 at 21:45
$begingroup$
Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
$endgroup$
– EdOverflow
Jan 7 at 21:45
$begingroup$
Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
$endgroup$
– Emilio Novati
Jan 7 at 21:51
$begingroup$
Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
$endgroup$
– Emilio Novati
Jan 7 at 21:51
$begingroup$
no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
$endgroup$
– Student123
Jan 7 at 21:59
$begingroup$
no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
$endgroup$
– Student123
Jan 7 at 21:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Assuming I understand the question correctly, we have:
$$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
$$therefore z = w^6 = 24.00534-2.52306i$$
answered Jan 7 at 21:35
Peter ForemanPeter Foreman
4138
$endgroup$
add a comment |
$begingroup$
I think you're close.
If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.
However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).
You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.
Do you see how to fix it?
That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.
answered Jan 7 at 21:35
JohnJohn
22.7k32450
$endgroup$
$begingroup$
I assume z=0.824-1.487i maybe?
$endgroup$
– Student123
Jan 7 at 21:44
$begingroup$
@Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
$endgroup$
– John Doe
Jan 7 at 22:05
$begingroup$
is it $z = w^6 = 24.00534-2.52306i$ ?
$endgroup$
– Student123
Jan 7 at 22:39
add a comment |
$begingroup$
Change the $+$ to a $-$ and you're golden.
Presumably you have calculated
$$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.
answered Jan 7 at 21:34
bouncebackbounceback
28218
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming I understand the question correctly, we have:
$$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
$$therefore z = w^6 = 24.00534-2.52306i$$
answered Jan 7 at 21:35
Peter ForemanPeter Foreman
4138
$endgroup$
add a comment |
$begingroup$
Assuming I understand the question correctly, we have:
$$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
$$therefore z = w^6 = 24.00534-2.52306i$$
answered Jan 7 at 21:35
Peter ForemanPeter Foreman
4138
$endgroup$
add a comment |
$begingroup$
Assuming I understand the question correctly, we have:
$$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
$$therefore z = w^6 = 24.00534-2.52306i$$
answered Jan 7 at 21:35
Peter ForemanPeter Foreman
4138
$endgroup$
Assuming I understand the question correctly, we have:
$$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
$$therefore z = w^6 = 24.00534-2.52306i$$
answered Jan 7 at 21:35
Peter ForemanPeter Foreman
4138
answered Jan 7 at 21:35
Peter ForemanPeter Foreman
4138
answered Jan 7 at 21:35
Peter ForemanPeter Foreman
4138
answered Jan 7 at 21:35
Peter ForemanPeter Foreman
4138
4138
add a comment |
add a comment |
$begingroup$
I think you're close.
If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.
However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).
You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.
Do you see how to fix it?
That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.
answered Jan 7 at 21:35
JohnJohn
22.7k32450
$endgroup$
$begingroup$
I assume z=0.824-1.487i maybe?
$endgroup$
– Student123
Jan 7 at 21:44
$begingroup$
@Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
$endgroup$
– John Doe
Jan 7 at 22:05
$begingroup$
is it $z = w^6 = 24.00534-2.52306i$ ?
$endgroup$
– Student123
Jan 7 at 22:39
add a comment |
$begingroup$
I think you're close.
If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.
However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).
You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.
Do you see how to fix it?
That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.
answered Jan 7 at 21:35
JohnJohn
22.7k32450
$endgroup$
$begingroup$
I assume z=0.824-1.487i maybe?
$endgroup$
– Student123
Jan 7 at 21:44
$begingroup$
@Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
$endgroup$
– John Doe
Jan 7 at 22:05
$begingroup$
is it $z = w^6 = 24.00534-2.52306i$ ?
$endgroup$
– Student123
Jan 7 at 22:39
add a comment |
$begingroup$
I think you're close.
If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.
However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).
You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.
Do you see how to fix it?
That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.
answered Jan 7 at 21:35
JohnJohn
22.7k32450
$endgroup$
I think you're close.
If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.
However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).
You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.
Do you see how to fix it?
That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.
answered Jan 7 at 21:35
JohnJohn
22.7k32450
answered Jan 7 at 21:35
JohnJohn
22.7k32450
answered Jan 7 at 21:35
JohnJohn
22.7k32450
answered Jan 7 at 21:35
JohnJohn
22.7k32450
22.7k32450
$begingroup$
I assume z=0.824-1.487i maybe?
$endgroup$
– Student123
Jan 7 at 21:44
$begingroup$
@Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
$endgroup$
– John Doe
Jan 7 at 22:05
$begingroup$
is it $z = w^6 = 24.00534-2.52306i$ ?
$endgroup$
– Student123
Jan 7 at 22:39
add a comment |
$begingroup$
I assume z=0.824-1.487i maybe?
$endgroup$
– Student123
Jan 7 at 21:44
$begingroup$
@Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
$endgroup$
– John Doe
Jan 7 at 22:05
$begingroup$
is it $z = w^6 = 24.00534-2.52306i$ ?
$endgroup$
– Student123
Jan 7 at 22:39
$begingroup$
I assume z=0.824-1.487i maybe?
$endgroup$
– Student123
Jan 7 at 21:44
$begingroup$
I assume z=0.824-1.487i maybe?
$endgroup$
– Student123
Jan 7 at 21:44
$begingroup$
@Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
$endgroup$
– John Doe
Jan 7 at 22:05
$begingroup$
@Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
$endgroup$
– John Doe
Jan 7 at 22:05
$begingroup$
is it $z = w^6 = 24.00534-2.52306i$ ?
$endgroup$
– Student123
Jan 7 at 22:39
$begingroup$
is it $z = w^6 = 24.00534-2.52306i$ ?
$endgroup$
– Student123
Jan 7 at 22:39
add a comment |
$begingroup$
Change the $+$ to a $-$ and you're golden.
Presumably you have calculated
$$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.
answered Jan 7 at 21:34
bouncebackbounceback
28218
$endgroup$
add a comment |
$begingroup$
Change the $+$ to a $-$ and you're golden.
Presumably you have calculated
$$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.
answered Jan 7 at 21:34
bouncebackbounceback
28218
$endgroup$
add a comment |
$begingroup$
Change the $+$ to a $-$ and you're golden.
Presumably you have calculated
$$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.
answered Jan 7 at 21:34
bouncebackbounceback
28218
$endgroup$
Change the $+$ to a $-$ and you're golden.
Presumably you have calculated
$$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.
answered Jan 7 at 21:34
bouncebackbounceback
28218
answered Jan 7 at 21:34
bouncebackbounceback
28218
answered Jan 7 at 21:34
bouncebackbounceback
28218
answered Jan 7 at 21:34
bouncebackbounceback
28218
28218
add a comment |
add a comment |
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$begingroup$
Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
$endgroup$
– EdOverflow
Jan 7 at 21:45
$begingroup$
Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
$endgroup$
– Emilio Novati
Jan 7 at 21:51
$begingroup$
no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
$endgroup$
– Student123
Jan 7 at 21:59