Why is real part written first in complex numbers?
$begingroup$
While expressing complex numbers as $a + iota b$, is there a specific reason for writing the real part before the imaginary part?
Who introduced this notation first? Is it a case where it just hung up with us due to the guy you introduced us to this?
soft-question notation
$endgroup$
add a comment |
$begingroup$
While expressing complex numbers as $a + iota b$, is there a specific reason for writing the real part before the imaginary part?
Who introduced this notation first? Is it a case where it just hung up with us due to the guy you introduced us to this?
soft-question notation
$endgroup$
1
$begingroup$
The addition in the field of complex numbers is commutative, as is multiplication. In other words $a+bi$, $bi+a$, $a+ib$ and $ib+a$ are all equal. Who says only one way of writing it is acceptable?
$endgroup$
– Jyrki Lahtonen
May 8 '14 at 6:41
2
$begingroup$
@JyrkiLahtonen It is not about it being acceptable. Only one way seems to be prevalent. In all of academia, that usually happens if some popular book which pioneered the topic introduced the notation. I would like to know the history.
$endgroup$
– Cheeku
May 8 '14 at 6:45
$begingroup$
No specific reason.
$endgroup$
– evil999man
May 8 '14 at 7:27
$begingroup$
For the same reason that we write the $x$-coordinate and then the $y$-coordinate when we write an ordered pair $(a,b)$.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 6:29
add a comment |
$begingroup$
While expressing complex numbers as $a + iota b$, is there a specific reason for writing the real part before the imaginary part?
Who introduced this notation first? Is it a case where it just hung up with us due to the guy you introduced us to this?
soft-question notation
$endgroup$
While expressing complex numbers as $a + iota b$, is there a specific reason for writing the real part before the imaginary part?
Who introduced this notation first? Is it a case where it just hung up with us due to the guy you introduced us to this?
soft-question notation
soft-question notation
asked May 8 '14 at 6:31
CheekuCheeku
601515
601515
1
$begingroup$
The addition in the field of complex numbers is commutative, as is multiplication. In other words $a+bi$, $bi+a$, $a+ib$ and $ib+a$ are all equal. Who says only one way of writing it is acceptable?
$endgroup$
– Jyrki Lahtonen
May 8 '14 at 6:41
2
$begingroup$
@JyrkiLahtonen It is not about it being acceptable. Only one way seems to be prevalent. In all of academia, that usually happens if some popular book which pioneered the topic introduced the notation. I would like to know the history.
$endgroup$
– Cheeku
May 8 '14 at 6:45
$begingroup$
No specific reason.
$endgroup$
– evil999man
May 8 '14 at 7:27
$begingroup$
For the same reason that we write the $x$-coordinate and then the $y$-coordinate when we write an ordered pair $(a,b)$.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 6:29
add a comment |
1
$begingroup$
The addition in the field of complex numbers is commutative, as is multiplication. In other words $a+bi$, $bi+a$, $a+ib$ and $ib+a$ are all equal. Who says only one way of writing it is acceptable?
$endgroup$
– Jyrki Lahtonen
May 8 '14 at 6:41
2
$begingroup$
@JyrkiLahtonen It is not about it being acceptable. Only one way seems to be prevalent. In all of academia, that usually happens if some popular book which pioneered the topic introduced the notation. I would like to know the history.
$endgroup$
– Cheeku
May 8 '14 at 6:45
$begingroup$
No specific reason.
$endgroup$
– evil999man
May 8 '14 at 7:27
$begingroup$
For the same reason that we write the $x$-coordinate and then the $y$-coordinate when we write an ordered pair $(a,b)$.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 6:29
1
1
$begingroup$
The addition in the field of complex numbers is commutative, as is multiplication. In other words $a+bi$, $bi+a$, $a+ib$ and $ib+a$ are all equal. Who says only one way of writing it is acceptable?
$endgroup$
– Jyrki Lahtonen
May 8 '14 at 6:41
$begingroup$
The addition in the field of complex numbers is commutative, as is multiplication. In other words $a+bi$, $bi+a$, $a+ib$ and $ib+a$ are all equal. Who says only one way of writing it is acceptable?
$endgroup$
– Jyrki Lahtonen
May 8 '14 at 6:41
2
2
$begingroup$
@JyrkiLahtonen It is not about it being acceptable. Only one way seems to be prevalent. In all of academia, that usually happens if some popular book which pioneered the topic introduced the notation. I would like to know the history.
$endgroup$
– Cheeku
May 8 '14 at 6:45
$begingroup$
@JyrkiLahtonen It is not about it being acceptable. Only one way seems to be prevalent. In all of academia, that usually happens if some popular book which pioneered the topic introduced the notation. I would like to know the history.
$endgroup$
– Cheeku
May 8 '14 at 6:45
$begingroup$
No specific reason.
$endgroup$
– evil999man
May 8 '14 at 7:27
$begingroup$
No specific reason.
$endgroup$
– evil999man
May 8 '14 at 7:27
$begingroup$
For the same reason that we write the $x$-coordinate and then the $y$-coordinate when we write an ordered pair $(a,b)$.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 6:29
$begingroup$
For the same reason that we write the $x$-coordinate and then the $y$-coordinate when we write an ordered pair $(a,b)$.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 6:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is a historical convention, and Euler himself chose it:
But there are some pretty good mathematical reasons to prefer it, which might explain why Euler chose it.
$
deflfrac#1#2{{largefrac{#1}{#2}}}
defrr{mathbb{R}}
$
At first, it might seem more natural to write $ai+b$, like "$ax+b$" where $x$ is the variable, but that is mostly useful only when de-emphasizing the constant terms. Here it is not the case, because $i$ is the constant. ${1,i}$ form the standard basis of the complex numbers over the reals, so in this setting one can think of a complex number as $a·1+b·i$, and we drop the "$1$" because it is redundant. If this was all, then there would be no reason to prefer one ordering of the basis elements over the other. However, $1$ is in fact very special, as it is the multiplicative identity.
This specialness of $1$ results in various facts that favour putting it first:
The reals embed into the complex numbers as numbers of the form $a+bi$ where $b = 0$. Thus it makes sense to put the real part first. The reals form a field, while the purely imaginary numbers do not, so we can see the complex numbers as a field extension of the reals, whereby it is natural to likewise extend the representation by appending a "${}+bi$".
$exp(iz) = cos(z) + i·sin(z)$ for any complex $z$. Obtained from the Taylor series, $cos(z)$ is the asymptotically significant term approximating $exp(iz)$ as $z to 0$. So it makes sense to put it first. Note also that we usually write "$i·sin(z)$" to emphasize the viewpoint that this equation relates $exp,cos,sin$, in which $i$ is merely a coefficient. Typographical issues established this form in history as $operatorname{cis}$, but I think it is very likely that anyone who rediscovers the identity will choose the same form.
Related to the above, $r·exp(it)$ for real $r,t$ is not only the polar form of a complex number but also can be seen as a spiral transform (scaling plus rotation) about the origin. The identity spiral transform is just $1 = cos(0)+i·sin(0)$, so again it makes sense to have the $cos(0)$ first.
When defining the complex logarithm as the inverse of the complex exponential function, namely via $ln(r·exp(it)) := ln(r)+it$ for real $r,t$ where $r > 0$ and $t in (-π,π]$, it makes sense to put the scaling $r$ first before the rotation $exp(it)$, because scalings are in some sense simpler. For example they preserve the reals, and have diagonal transformation matrices.
$endgroup$
1
$begingroup$
In what sense is $cos z$ the first-order term? I would call it $0$th or second-order, and $isin z$ first-order.
$endgroup$
– mr_e_man
Dec 18 '18 at 0:43
$begingroup$
@mr_e_man: I have always thought that English meaning of "first-order" in approximation is that it is the most significant. When comparing $cos(z)$ and $i·sin(z)$, the most significant is $cos(z)$. But to remove ambiguity, I'll edit my answer. Thanks! =)
$endgroup$
– user21820
Dec 18 '18 at 3:17
add a comment |
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$begingroup$
It is a historical convention, and Euler himself chose it:
But there are some pretty good mathematical reasons to prefer it, which might explain why Euler chose it.
$
deflfrac#1#2{{largefrac{#1}{#2}}}
defrr{mathbb{R}}
$
At first, it might seem more natural to write $ai+b$, like "$ax+b$" where $x$ is the variable, but that is mostly useful only when de-emphasizing the constant terms. Here it is not the case, because $i$ is the constant. ${1,i}$ form the standard basis of the complex numbers over the reals, so in this setting one can think of a complex number as $a·1+b·i$, and we drop the "$1$" because it is redundant. If this was all, then there would be no reason to prefer one ordering of the basis elements over the other. However, $1$ is in fact very special, as it is the multiplicative identity.
This specialness of $1$ results in various facts that favour putting it first:
The reals embed into the complex numbers as numbers of the form $a+bi$ where $b = 0$. Thus it makes sense to put the real part first. The reals form a field, while the purely imaginary numbers do not, so we can see the complex numbers as a field extension of the reals, whereby it is natural to likewise extend the representation by appending a "${}+bi$".
$exp(iz) = cos(z) + i·sin(z)$ for any complex $z$. Obtained from the Taylor series, $cos(z)$ is the asymptotically significant term approximating $exp(iz)$ as $z to 0$. So it makes sense to put it first. Note also that we usually write "$i·sin(z)$" to emphasize the viewpoint that this equation relates $exp,cos,sin$, in which $i$ is merely a coefficient. Typographical issues established this form in history as $operatorname{cis}$, but I think it is very likely that anyone who rediscovers the identity will choose the same form.
Related to the above, $r·exp(it)$ for real $r,t$ is not only the polar form of a complex number but also can be seen as a spiral transform (scaling plus rotation) about the origin. The identity spiral transform is just $1 = cos(0)+i·sin(0)$, so again it makes sense to have the $cos(0)$ first.
When defining the complex logarithm as the inverse of the complex exponential function, namely via $ln(r·exp(it)) := ln(r)+it$ for real $r,t$ where $r > 0$ and $t in (-π,π]$, it makes sense to put the scaling $r$ first before the rotation $exp(it)$, because scalings are in some sense simpler. For example they preserve the reals, and have diagonal transformation matrices.
$endgroup$
1
$begingroup$
In what sense is $cos z$ the first-order term? I would call it $0$th or second-order, and $isin z$ first-order.
$endgroup$
– mr_e_man
Dec 18 '18 at 0:43
$begingroup$
@mr_e_man: I have always thought that English meaning of "first-order" in approximation is that it is the most significant. When comparing $cos(z)$ and $i·sin(z)$, the most significant is $cos(z)$. But to remove ambiguity, I'll edit my answer. Thanks! =)
$endgroup$
– user21820
Dec 18 '18 at 3:17
add a comment |
$begingroup$
It is a historical convention, and Euler himself chose it:
But there are some pretty good mathematical reasons to prefer it, which might explain why Euler chose it.
$
deflfrac#1#2{{largefrac{#1}{#2}}}
defrr{mathbb{R}}
$
At first, it might seem more natural to write $ai+b$, like "$ax+b$" where $x$ is the variable, but that is mostly useful only when de-emphasizing the constant terms. Here it is not the case, because $i$ is the constant. ${1,i}$ form the standard basis of the complex numbers over the reals, so in this setting one can think of a complex number as $a·1+b·i$, and we drop the "$1$" because it is redundant. If this was all, then there would be no reason to prefer one ordering of the basis elements over the other. However, $1$ is in fact very special, as it is the multiplicative identity.
This specialness of $1$ results in various facts that favour putting it first:
The reals embed into the complex numbers as numbers of the form $a+bi$ where $b = 0$. Thus it makes sense to put the real part first. The reals form a field, while the purely imaginary numbers do not, so we can see the complex numbers as a field extension of the reals, whereby it is natural to likewise extend the representation by appending a "${}+bi$".
$exp(iz) = cos(z) + i·sin(z)$ for any complex $z$. Obtained from the Taylor series, $cos(z)$ is the asymptotically significant term approximating $exp(iz)$ as $z to 0$. So it makes sense to put it first. Note also that we usually write "$i·sin(z)$" to emphasize the viewpoint that this equation relates $exp,cos,sin$, in which $i$ is merely a coefficient. Typographical issues established this form in history as $operatorname{cis}$, but I think it is very likely that anyone who rediscovers the identity will choose the same form.
Related to the above, $r·exp(it)$ for real $r,t$ is not only the polar form of a complex number but also can be seen as a spiral transform (scaling plus rotation) about the origin. The identity spiral transform is just $1 = cos(0)+i·sin(0)$, so again it makes sense to have the $cos(0)$ first.
When defining the complex logarithm as the inverse of the complex exponential function, namely via $ln(r·exp(it)) := ln(r)+it$ for real $r,t$ where $r > 0$ and $t in (-π,π]$, it makes sense to put the scaling $r$ first before the rotation $exp(it)$, because scalings are in some sense simpler. For example they preserve the reals, and have diagonal transformation matrices.
$endgroup$
1
$begingroup$
In what sense is $cos z$ the first-order term? I would call it $0$th or second-order, and $isin z$ first-order.
$endgroup$
– mr_e_man
Dec 18 '18 at 0:43
$begingroup$
@mr_e_man: I have always thought that English meaning of "first-order" in approximation is that it is the most significant. When comparing $cos(z)$ and $i·sin(z)$, the most significant is $cos(z)$. But to remove ambiguity, I'll edit my answer. Thanks! =)
$endgroup$
– user21820
Dec 18 '18 at 3:17
add a comment |
$begingroup$
It is a historical convention, and Euler himself chose it:
But there are some pretty good mathematical reasons to prefer it, which might explain why Euler chose it.
$
deflfrac#1#2{{largefrac{#1}{#2}}}
defrr{mathbb{R}}
$
At first, it might seem more natural to write $ai+b$, like "$ax+b$" where $x$ is the variable, but that is mostly useful only when de-emphasizing the constant terms. Here it is not the case, because $i$ is the constant. ${1,i}$ form the standard basis of the complex numbers over the reals, so in this setting one can think of a complex number as $a·1+b·i$, and we drop the "$1$" because it is redundant. If this was all, then there would be no reason to prefer one ordering of the basis elements over the other. However, $1$ is in fact very special, as it is the multiplicative identity.
This specialness of $1$ results in various facts that favour putting it first:
The reals embed into the complex numbers as numbers of the form $a+bi$ where $b = 0$. Thus it makes sense to put the real part first. The reals form a field, while the purely imaginary numbers do not, so we can see the complex numbers as a field extension of the reals, whereby it is natural to likewise extend the representation by appending a "${}+bi$".
$exp(iz) = cos(z) + i·sin(z)$ for any complex $z$. Obtained from the Taylor series, $cos(z)$ is the asymptotically significant term approximating $exp(iz)$ as $z to 0$. So it makes sense to put it first. Note also that we usually write "$i·sin(z)$" to emphasize the viewpoint that this equation relates $exp,cos,sin$, in which $i$ is merely a coefficient. Typographical issues established this form in history as $operatorname{cis}$, but I think it is very likely that anyone who rediscovers the identity will choose the same form.
Related to the above, $r·exp(it)$ for real $r,t$ is not only the polar form of a complex number but also can be seen as a spiral transform (scaling plus rotation) about the origin. The identity spiral transform is just $1 = cos(0)+i·sin(0)$, so again it makes sense to have the $cos(0)$ first.
When defining the complex logarithm as the inverse of the complex exponential function, namely via $ln(r·exp(it)) := ln(r)+it$ for real $r,t$ where $r > 0$ and $t in (-π,π]$, it makes sense to put the scaling $r$ first before the rotation $exp(it)$, because scalings are in some sense simpler. For example they preserve the reals, and have diagonal transformation matrices.
$endgroup$
It is a historical convention, and Euler himself chose it:
But there are some pretty good mathematical reasons to prefer it, which might explain why Euler chose it.
$
deflfrac#1#2{{largefrac{#1}{#2}}}
defrr{mathbb{R}}
$
At first, it might seem more natural to write $ai+b$, like "$ax+b$" where $x$ is the variable, but that is mostly useful only when de-emphasizing the constant terms. Here it is not the case, because $i$ is the constant. ${1,i}$ form the standard basis of the complex numbers over the reals, so in this setting one can think of a complex number as $a·1+b·i$, and we drop the "$1$" because it is redundant. If this was all, then there would be no reason to prefer one ordering of the basis elements over the other. However, $1$ is in fact very special, as it is the multiplicative identity.
This specialness of $1$ results in various facts that favour putting it first:
The reals embed into the complex numbers as numbers of the form $a+bi$ where $b = 0$. Thus it makes sense to put the real part first. The reals form a field, while the purely imaginary numbers do not, so we can see the complex numbers as a field extension of the reals, whereby it is natural to likewise extend the representation by appending a "${}+bi$".
$exp(iz) = cos(z) + i·sin(z)$ for any complex $z$. Obtained from the Taylor series, $cos(z)$ is the asymptotically significant term approximating $exp(iz)$ as $z to 0$. So it makes sense to put it first. Note also that we usually write "$i·sin(z)$" to emphasize the viewpoint that this equation relates $exp,cos,sin$, in which $i$ is merely a coefficient. Typographical issues established this form in history as $operatorname{cis}$, but I think it is very likely that anyone who rediscovers the identity will choose the same form.
Related to the above, $r·exp(it)$ for real $r,t$ is not only the polar form of a complex number but also can be seen as a spiral transform (scaling plus rotation) about the origin. The identity spiral transform is just $1 = cos(0)+i·sin(0)$, so again it makes sense to have the $cos(0)$ first.
When defining the complex logarithm as the inverse of the complex exponential function, namely via $ln(r·exp(it)) := ln(r)+it$ for real $r,t$ where $r > 0$ and $t in (-π,π]$, it makes sense to put the scaling $r$ first before the rotation $exp(it)$, because scalings are in some sense simpler. For example they preserve the reals, and have diagonal transformation matrices.
edited Dec 18 '18 at 3:18
answered Dec 17 '18 at 7:33
user21820user21820
39.4k543155
39.4k543155
1
$begingroup$
In what sense is $cos z$ the first-order term? I would call it $0$th or second-order, and $isin z$ first-order.
$endgroup$
– mr_e_man
Dec 18 '18 at 0:43
$begingroup$
@mr_e_man: I have always thought that English meaning of "first-order" in approximation is that it is the most significant. When comparing $cos(z)$ and $i·sin(z)$, the most significant is $cos(z)$. But to remove ambiguity, I'll edit my answer. Thanks! =)
$endgroup$
– user21820
Dec 18 '18 at 3:17
add a comment |
1
$begingroup$
In what sense is $cos z$ the first-order term? I would call it $0$th or second-order, and $isin z$ first-order.
$endgroup$
– mr_e_man
Dec 18 '18 at 0:43
$begingroup$
@mr_e_man: I have always thought that English meaning of "first-order" in approximation is that it is the most significant. When comparing $cos(z)$ and $i·sin(z)$, the most significant is $cos(z)$. But to remove ambiguity, I'll edit my answer. Thanks! =)
$endgroup$
– user21820
Dec 18 '18 at 3:17
1
1
$begingroup$
In what sense is $cos z$ the first-order term? I would call it $0$th or second-order, and $isin z$ first-order.
$endgroup$
– mr_e_man
Dec 18 '18 at 0:43
$begingroup$
In what sense is $cos z$ the first-order term? I would call it $0$th or second-order, and $isin z$ first-order.
$endgroup$
– mr_e_man
Dec 18 '18 at 0:43
$begingroup$
@mr_e_man: I have always thought that English meaning of "first-order" in approximation is that it is the most significant. When comparing $cos(z)$ and $i·sin(z)$, the most significant is $cos(z)$. But to remove ambiguity, I'll edit my answer. Thanks! =)
$endgroup$
– user21820
Dec 18 '18 at 3:17
$begingroup$
@mr_e_man: I have always thought that English meaning of "first-order" in approximation is that it is the most significant. When comparing $cos(z)$ and $i·sin(z)$, the most significant is $cos(z)$. But to remove ambiguity, I'll edit my answer. Thanks! =)
$endgroup$
– user21820
Dec 18 '18 at 3:17
add a comment |
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$begingroup$
The addition in the field of complex numbers is commutative, as is multiplication. In other words $a+bi$, $bi+a$, $a+ib$ and $ib+a$ are all equal. Who says only one way of writing it is acceptable?
$endgroup$
– Jyrki Lahtonen
May 8 '14 at 6:41
2
$begingroup$
@JyrkiLahtonen It is not about it being acceptable. Only one way seems to be prevalent. In all of academia, that usually happens if some popular book which pioneered the topic introduced the notation. I would like to know the history.
$endgroup$
– Cheeku
May 8 '14 at 6:45
$begingroup$
No specific reason.
$endgroup$
– evil999man
May 8 '14 at 7:27
$begingroup$
For the same reason that we write the $x$-coordinate and then the $y$-coordinate when we write an ordered pair $(a,b)$.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 6:29