Mixing
Negative modulus in complex numbers?
$begingroup$
This is an introducory task from an exam.
If $z = -2(cos{5} - isin{5})$, then what are:
$Re(z), Im(z), arg(z)$ and $ |z|$?
First of all, how is it possible that the modulus is negative $|z|=-2$? Or is the modulus actually $|z|= 2$ and the minus is kind of in front of everything, and that's why the sign inside of the brackets is changed as well? That would make some sense.
I assume $arg(z) = 5$. How do I calculate $Re(z) $ and $Im(z)$? Something like this should do the job?
$$arg(z) = frac{Re(z)}{|z|}$$
$$5 = frac{Re(z)}{2}$$
$$10 = Re(z)$$
And analogically with $Im(z):$
$$arg(z) = frac{Im(z)}{|z|}$$
$$5 = frac{Im(z)}{2} Rightarrow Im(z) = Re(z) = 10$$
I'm sure I'm confusing something here because, probably somewhere wrong $pm$ signs.
Help's appreciated.
And finally: is there some good calculator for complex numbers? Let's say I have a polar form and I want to find out the $Re(z), Im(z)$ and such. Wolframalpha seems like doesn't work fine for that.
complex-numbers
edited Jan 30 at 15:28
José Carlos Santos
172k22132239
asked Jan 30 at 15:21
wenoweno
37011
$endgroup$
add a comment |
$begingroup$
This is an introducory task from an exam.
If $z = -2(cos{5} - isin{5})$, then what are:
$Re(z), Im(z), arg(z)$ and $ |z|$?
First of all, how is it possible that the modulus is negative $|z|=-2$? Or is the modulus actually $|z|= 2$ and the minus is kind of in front of everything, and that's why the sign inside of the brackets is changed as well? That would make some sense.
I assume $arg(z) = 5$. How do I calculate $Re(z) $ and $Im(z)$? Something like this should do the job?
$$arg(z) = frac{Re(z)}{|z|}$$
$$5 = frac{Re(z)}{2}$$
$$10 = Re(z)$$
And analogically with $Im(z):$
$$arg(z) = frac{Im(z)}{|z|}$$
$$5 = frac{Im(z)}{2} Rightarrow Im(z) = Re(z) = 10$$
I'm sure I'm confusing something here because, probably somewhere wrong $pm$ signs.
Help's appreciated.
And finally: is there some good calculator for complex numbers? Let's say I have a polar form and I want to find out the $Re(z), Im(z)$ and such. Wolframalpha seems like doesn't work fine for that.
complex-numbers
edited Jan 30 at 15:28
José Carlos Santos
172k22132239
asked Jan 30 at 15:21
wenoweno
37011
$endgroup$
1
$begingroup$
The modulus of a complex number can't be negative
$endgroup$
– J. W. Tanner
Jan 30 at 15:24
2
$begingroup$
From what you've said, you are trying to solve the problem while having no idea what "arg" "re" and "Im" are. That is your problem. Fix that first. Randomly flailing around is no substitute for looking up what they actually mean. Several things you wrote definitely fall into the "flailing" category.
$endgroup$
– rschwieb
Jan 30 at 15:29
add a comment |
$begingroup$
This is an introducory task from an exam.
If $z = -2(cos{5} - isin{5})$, then what are:
$Re(z), Im(z), arg(z)$ and $ |z|$?
First of all, how is it possible that the modulus is negative $|z|=-2$? Or is the modulus actually $|z|= 2$ and the minus is kind of in front of everything, and that's why the sign inside of the brackets is changed as well? That would make some sense.
I assume $arg(z) = 5$. How do I calculate $Re(z) $ and $Im(z)$? Something like this should do the job?
$$arg(z) = frac{Re(z)}{|z|}$$
$$5 = frac{Re(z)}{2}$$
$$10 = Re(z)$$
And analogically with $Im(z):$
$$arg(z) = frac{Im(z)}{|z|}$$
$$5 = frac{Im(z)}{2} Rightarrow Im(z) = Re(z) = 10$$
I'm sure I'm confusing something here because, probably somewhere wrong $pm$ signs.
Help's appreciated.
And finally: is there some good calculator for complex numbers? Let's say I have a polar form and I want to find out the $Re(z), Im(z)$ and such. Wolframalpha seems like doesn't work fine for that.
complex-numbers
edited Jan 30 at 15:28
José Carlos Santos
172k22132239
asked Jan 30 at 15:21
wenoweno
37011
$endgroup$
This is an introducory task from an exam.
If $z = -2(cos{5} - isin{5})$, then what are:
$Re(z), Im(z), arg(z)$ and $ |z|$?
First of all, how is it possible that the modulus is negative $|z|=-2$? Or is the modulus actually $|z|= 2$ and the minus is kind of in front of everything, and that's why the sign inside of the brackets is changed as well? That would make some sense.
I assume $arg(z) = 5$. How do I calculate $Re(z) $ and $Im(z)$? Something like this should do the job?
$$arg(z) = frac{Re(z)}{|z|}$$
$$5 = frac{Re(z)}{2}$$
$$10 = Re(z)$$
And analogically with $Im(z):$
$$arg(z) = frac{Im(z)}{|z|}$$
$$5 = frac{Im(z)}{2} Rightarrow Im(z) = Re(z) = 10$$
I'm sure I'm confusing something here because, probably somewhere wrong $pm$ signs.
Help's appreciated.
And finally: is there some good calculator for complex numbers? Let's say I have a polar form and I want to find out the $Re(z), Im(z)$ and such. Wolframalpha seems like doesn't work fine for that.
complex-numbers
complex-numbers
edited Jan 30 at 15:28
José Carlos Santos
172k22132239
asked Jan 30 at 15:21
wenoweno
37011
edited Jan 30 at 15:28
José Carlos Santos
172k22132239
asked Jan 30 at 15:21
wenoweno
37011
edited Jan 30 at 15:28
José Carlos Santos
172k22132239
edited Jan 30 at 15:28
José Carlos Santos
172k22132239
edited Jan 30 at 15:28
José Carlos Santos
172k22132239
172k22132239
asked Jan 30 at 15:21
wenoweno
37011
asked Jan 30 at 15:21
wenoweno
37011
asked Jan 30 at 15:21
wenoweno
37011
37011
1
$begingroup$
The modulus of a complex number can't be negative
$endgroup$
– J. W. Tanner
Jan 30 at 15:24
2
$begingroup$
From what you've said, you are trying to solve the problem while having no idea what "arg" "re" and "Im" are. That is your problem. Fix that first. Randomly flailing around is no substitute for looking up what they actually mean. Several things you wrote definitely fall into the "flailing" category.
$endgroup$
– rschwieb
Jan 30 at 15:29
add a comment |
1
$begingroup$
The modulus of a complex number can't be negative
$endgroup$
– J. W. Tanner
Jan 30 at 15:24
2
$begingroup$
From what you've said, you are trying to solve the problem while having no idea what "arg" "re" and "Im" are. That is your problem. Fix that first. Randomly flailing around is no substitute for looking up what they actually mean. Several things you wrote definitely fall into the "flailing" category.
$endgroup$
– rschwieb
Jan 30 at 15:29
1
1
$begingroup$
The modulus of a complex number can't be negative
$endgroup$
– J. W. Tanner
Jan 30 at 15:24
$begingroup$
The modulus of a complex number can't be negative
$endgroup$
– J. W. Tanner
Jan 30 at 15:24
2
2
$begingroup$
From what you've said, you are trying to solve the problem while having no idea what "arg" "re" and "Im" are. That is your problem. Fix that first. Randomly flailing around is no substitute for looking up what they actually mean. Several things you wrote definitely fall into the "flailing" category.
$endgroup$
– rschwieb
Jan 30 at 15:29
$begingroup$
From what you've said, you are trying to solve the problem while having no idea what "arg" "re" and "Im" are. That is your problem. Fix that first. Randomly flailing around is no substitute for looking up what they actually mean. Several things you wrote definitely fall into the "flailing" category.
$endgroup$
– rschwieb
Jan 30 at 15:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Currently, the number is not in polar form, as it should be in the form $r(cos(theta) + i sin(theta))$, where $r ge 0$. Note the $+$ sign, and the non-negative number $r$ out the front. Every complex number, including the one given, has a polar form (in fact, infinitely many), and from this you can read off the modulus and argument. But, since this is not in polar form, you need to do some extra work.
First, try absorbing the minus sign into the brackets:
$$2(-cos 5 + i sin 5).$$
Then, recalling that $sin(pi - x) = sin(x)$ and $cos(pi - x) = -cos(x)$, we get
$$2(cos(pi - 5) + i sin(pi - 5)).$$
This is now in polar form. The modulus is $2$, and one of the infinitely many arguments is $pi - 5$.
answered Jan 30 at 15:29
Theo BenditTheo Bendit
20.2k12353
$endgroup$
$begingroup$
Clever, thanks. Is there a way without modifying the arguments of $sin$ and $cos$? That is: is there a way of solving this by operating on $pm cos{5} land pm sin{5}$? Without using the $sin(pi - x) $ and $cos{pi - x}$ formulas?
$endgroup$
– weno
Jan 30 at 15:32
1
$begingroup$
@weno You could avoid the reflection formulas temporarily by looking at the $tan(operatorname{arg}(x + iy)) = y/x$ formula, remembering not just to take $arctan$ of both sides, but also to check which quadrant the complex number lies in. This will be a little messy though, and remember, the only possible arguments are of the form $(2k + 1)pi - 5$, so at some point, there's going to have to be some reflection somewhere.
$endgroup$
– Theo Bendit
Jan 30 at 15:37
add a comment |
$begingroup$
If $z=a+bi$, with $a,binmathbb R$, then $lvert zrvert=sqrt{a^2+b^2}$; in particular, $lvert zrvertgeqslant0$ for any $zinmathbb C$.
Actually, $bigllvert-2(cos5-isin5)bigrrvert=2$, $operatorname{Re}bigl(-2(cos5-isin5)bigr)=-2cos5$, and $operatorname{Im}bigl(-2(cos5-isin5)bigr)=2sin5$.
answered Jan 30 at 15:25
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Sorry but only one answer can be accepted. Thank you though. Directly finding $Re$, $Im$ seems elegant.
$endgroup$
– weno
Jan 30 at 15:43
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
Currently, the number is not in polar form, as it should be in the form $r(cos(theta) + i sin(theta))$, where $r ge 0$. Note the $+$ sign, and the non-negative number $r$ out the front. Every complex number, including the one given, has a polar form (in fact, infinitely many), and from this you can read off the modulus and argument. But, since this is not in polar form, you need to do some extra work.
First, try absorbing the minus sign into the brackets:
$$2(-cos 5 + i sin 5).$$
Then, recalling that $sin(pi - x) = sin(x)$ and $cos(pi - x) = -cos(x)$, we get
$$2(cos(pi - 5) + i sin(pi - 5)).$$
This is now in polar form. The modulus is $2$, and one of the infinitely many arguments is $pi - 5$.
answered Jan 30 at 15:29
Theo BenditTheo Bendit
20.2k12353
$endgroup$
$begingroup$
Clever, thanks. Is there a way without modifying the arguments of $sin$ and $cos$? That is: is there a way of solving this by operating on $pm cos{5} land pm sin{5}$? Without using the $sin(pi - x) $ and $cos{pi - x}$ formulas?
$endgroup$
– weno
Jan 30 at 15:32
1
$begingroup$
@weno You could avoid the reflection formulas temporarily by looking at the $tan(operatorname{arg}(x + iy)) = y/x$ formula, remembering not just to take $arctan$ of both sides, but also to check which quadrant the complex number lies in. This will be a little messy though, and remember, the only possible arguments are of the form $(2k + 1)pi - 5$, so at some point, there's going to have to be some reflection somewhere.
$endgroup$
– Theo Bendit
Jan 30 at 15:37
add a comment |
$begingroup$
Currently, the number is not in polar form, as it should be in the form $r(cos(theta) + i sin(theta))$, where $r ge 0$. Note the $+$ sign, and the non-negative number $r$ out the front. Every complex number, including the one given, has a polar form (in fact, infinitely many), and from this you can read off the modulus and argument. But, since this is not in polar form, you need to do some extra work.
First, try absorbing the minus sign into the brackets:
$$2(-cos 5 + i sin 5).$$
Then, recalling that $sin(pi - x) = sin(x)$ and $cos(pi - x) = -cos(x)$, we get
$$2(cos(pi - 5) + i sin(pi - 5)).$$
This is now in polar form. The modulus is $2$, and one of the infinitely many arguments is $pi - 5$.
answered Jan 30 at 15:29
Theo BenditTheo Bendit
20.2k12353
$endgroup$
$begingroup$
Clever, thanks. Is there a way without modifying the arguments of $sin$ and $cos$? That is: is there a way of solving this by operating on $pm cos{5} land pm sin{5}$? Without using the $sin(pi - x) $ and $cos{pi - x}$ formulas?
$endgroup$
– weno
Jan 30 at 15:32
1
$begingroup$
@weno You could avoid the reflection formulas temporarily by looking at the $tan(operatorname{arg}(x + iy)) = y/x$ formula, remembering not just to take $arctan$ of both sides, but also to check which quadrant the complex number lies in. This will be a little messy though, and remember, the only possible arguments are of the form $(2k + 1)pi - 5$, so at some point, there's going to have to be some reflection somewhere.
$endgroup$
– Theo Bendit
Jan 30 at 15:37
add a comment |
$begingroup$
Currently, the number is not in polar form, as it should be in the form $r(cos(theta) + i sin(theta))$, where $r ge 0$. Note the $+$ sign, and the non-negative number $r$ out the front. Every complex number, including the one given, has a polar form (in fact, infinitely many), and from this you can read off the modulus and argument. But, since this is not in polar form, you need to do some extra work.
First, try absorbing the minus sign into the brackets:
$$2(-cos 5 + i sin 5).$$
Then, recalling that $sin(pi - x) = sin(x)$ and $cos(pi - x) = -cos(x)$, we get
$$2(cos(pi - 5) + i sin(pi - 5)).$$
This is now in polar form. The modulus is $2$, and one of the infinitely many arguments is $pi - 5$.
answered Jan 30 at 15:29
Theo BenditTheo Bendit
20.2k12353
$endgroup$
Currently, the number is not in polar form, as it should be in the form $r(cos(theta) + i sin(theta))$, where $r ge 0$. Note the $+$ sign, and the non-negative number $r$ out the front. Every complex number, including the one given, has a polar form (in fact, infinitely many), and from this you can read off the modulus and argument. But, since this is not in polar form, you need to do some extra work.
First, try absorbing the minus sign into the brackets:
$$2(-cos 5 + i sin 5).$$
Then, recalling that $sin(pi - x) = sin(x)$ and $cos(pi - x) = -cos(x)$, we get
$$2(cos(pi - 5) + i sin(pi - 5)).$$
This is now in polar form. The modulus is $2$, and one of the infinitely many arguments is $pi - 5$.
answered Jan 30 at 15:29
Theo BenditTheo Bendit
20.2k12353
answered Jan 30 at 15:29
Theo BenditTheo Bendit
20.2k12353
answered Jan 30 at 15:29
Theo BenditTheo Bendit
20.2k12353
answered Jan 30 at 15:29
Theo BenditTheo Bendit
20.2k12353
20.2k12353
$begingroup$
Clever, thanks. Is there a way without modifying the arguments of $sin$ and $cos$? That is: is there a way of solving this by operating on $pm cos{5} land pm sin{5}$? Without using the $sin(pi - x) $ and $cos{pi - x}$ formulas?
$endgroup$
– weno
Jan 30 at 15:32
1
$begingroup$
@weno You could avoid the reflection formulas temporarily by looking at the $tan(operatorname{arg}(x + iy)) = y/x$ formula, remembering not just to take $arctan$ of both sides, but also to check which quadrant the complex number lies in. This will be a little messy though, and remember, the only possible arguments are of the form $(2k + 1)pi - 5$, so at some point, there's going to have to be some reflection somewhere.
$endgroup$
– Theo Bendit
Jan 30 at 15:37
add a comment |
$begingroup$
Clever, thanks. Is there a way without modifying the arguments of $sin$ and $cos$? That is: is there a way of solving this by operating on $pm cos{5} land pm sin{5}$? Without using the $sin(pi - x) $ and $cos{pi - x}$ formulas?
$endgroup$
– weno
Jan 30 at 15:32
1
$begingroup$
@weno You could avoid the reflection formulas temporarily by looking at the $tan(operatorname{arg}(x + iy)) = y/x$ formula, remembering not just to take $arctan$ of both sides, but also to check which quadrant the complex number lies in. This will be a little messy though, and remember, the only possible arguments are of the form $(2k + 1)pi - 5$, so at some point, there's going to have to be some reflection somewhere.
$endgroup$
– Theo Bendit
Jan 30 at 15:37
$begingroup$
Clever, thanks. Is there a way without modifying the arguments of $sin$ and $cos$? That is: is there a way of solving this by operating on $pm cos{5} land pm sin{5}$? Without using the $sin(pi - x) $ and $cos{pi - x}$ formulas?
$endgroup$
– weno
Jan 30 at 15:32
$begingroup$
Clever, thanks. Is there a way without modifying the arguments of $sin$ and $cos$? That is: is there a way of solving this by operating on $pm cos{5} land pm sin{5}$? Without using the $sin(pi - x) $ and $cos{pi - x}$ formulas?
$endgroup$
– weno
Jan 30 at 15:32
1
1
$begingroup$
@weno You could avoid the reflection formulas temporarily by looking at the $tan(operatorname{arg}(x + iy)) = y/x$ formula, remembering not just to take $arctan$ of both sides, but also to check which quadrant the complex number lies in. This will be a little messy though, and remember, the only possible arguments are of the form $(2k + 1)pi - 5$, so at some point, there's going to have to be some reflection somewhere.
$endgroup$
– Theo Bendit
Jan 30 at 15:37
$begingroup$
@weno You could avoid the reflection formulas temporarily by looking at the $tan(operatorname{arg}(x + iy)) = y/x$ formula, remembering not just to take $arctan$ of both sides, but also to check which quadrant the complex number lies in. This will be a little messy though, and remember, the only possible arguments are of the form $(2k + 1)pi - 5$, so at some point, there's going to have to be some reflection somewhere.
$endgroup$
– Theo Bendit
Jan 30 at 15:37
add a comment |
$begingroup$
If $z=a+bi$, with $a,binmathbb R$, then $lvert zrvert=sqrt{a^2+b^2}$; in particular, $lvert zrvertgeqslant0$ for any $zinmathbb C$.
Actually, $bigllvert-2(cos5-isin5)bigrrvert=2$, $operatorname{Re}bigl(-2(cos5-isin5)bigr)=-2cos5$, and $operatorname{Im}bigl(-2(cos5-isin5)bigr)=2sin5$.
answered Jan 30 at 15:25
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Sorry but only one answer can be accepted. Thank you though. Directly finding $Re$, $Im$ seems elegant.
$endgroup$
– weno
Jan 30 at 15:43
add a comment |
$begingroup$
If $z=a+bi$, with $a,binmathbb R$, then $lvert zrvert=sqrt{a^2+b^2}$; in particular, $lvert zrvertgeqslant0$ for any $zinmathbb C$.
Actually, $bigllvert-2(cos5-isin5)bigrrvert=2$, $operatorname{Re}bigl(-2(cos5-isin5)bigr)=-2cos5$, and $operatorname{Im}bigl(-2(cos5-isin5)bigr)=2sin5$.
answered Jan 30 at 15:25
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Sorry but only one answer can be accepted. Thank you though. Directly finding $Re$, $Im$ seems elegant.
$endgroup$
– weno
Jan 30 at 15:43
add a comment |
$begingroup$
If $z=a+bi$, with $a,binmathbb R$, then $lvert zrvert=sqrt{a^2+b^2}$; in particular, $lvert zrvertgeqslant0$ for any $zinmathbb C$.
Actually, $bigllvert-2(cos5-isin5)bigrrvert=2$, $operatorname{Re}bigl(-2(cos5-isin5)bigr)=-2cos5$, and $operatorname{Im}bigl(-2(cos5-isin5)bigr)=2sin5$.
answered Jan 30 at 15:25
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
If $z=a+bi$, with $a,binmathbb R$, then $lvert zrvert=sqrt{a^2+b^2}$; in particular, $lvert zrvertgeqslant0$ for any $zinmathbb C$.
Actually, $bigllvert-2(cos5-isin5)bigrrvert=2$, $operatorname{Re}bigl(-2(cos5-isin5)bigr)=-2cos5$, and $operatorname{Im}bigl(-2(cos5-isin5)bigr)=2sin5$.
answered Jan 30 at 15:25
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 30 at 15:25
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 30 at 15:25
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 30 at 15:25
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
$begingroup$
Sorry but only one answer can be accepted. Thank you though. Directly finding $Re$, $Im$ seems elegant.
$endgroup$
– weno
Jan 30 at 15:43
add a comment |
$begingroup$
Sorry but only one answer can be accepted. Thank you though. Directly finding $Re$, $Im$ seems elegant.
$endgroup$
– weno
Jan 30 at 15:43
$begingroup$
Sorry but only one answer can be accepted. Thank you though. Directly finding $Re$, $Im$ seems elegant.
$endgroup$
– weno
Jan 30 at 15:43
$begingroup$
Sorry but only one answer can be accepted. Thank you though. Directly finding $Re$, $Im$ seems elegant.
$endgroup$
– weno
Jan 30 at 15:43
add a comment |
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$begingroup$
The modulus of a complex number can't be negative
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– J. W. Tanner
Jan 30 at 15:24
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From what you've said, you are trying to solve the problem while having no idea what "arg" "re" and "Im" are. That is your problem. Fix that first. Randomly flailing around is no substitute for looking up what they actually mean. Several things you wrote definitely fall into the "flailing" category.
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– rschwieb
Jan 30 at 15:29