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Is a negative logarithm meaningless?
Choosing the branch of a logarithmDetermination of complex logarithmWhy the complex logarithm function$ln(z)$ is not meromorphic on the whole complex planeBasic Logarithm equation, and how best to approach this question logicallyWhy does the integral $intfrac{1}{x+i}dx$ not require the absolute value in the logarithm?Riemann Zeta Function integralLimit to infinity and infinite logarithms?Simplifying Square Roots of a Negative NumberInverse trignometric functions in complex planeUnexpected examples of natural logarithm
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Is $log(-x)$, where $x in (0, infty)$ undefined? From solving quadratics I was first told that the discriminant has to be non-negative, since $sqrt{-a}$, where $a in (0, infty)$ is undefined. But this was before learning about Imaginary numbers and the complex plane.
For example, what would $log_3(-9)$ be? Is there a number $a$ such that $3^a=-9$?
So can you evaluate negative logarithms?
real-analysis complex-analysis logarithms
asked Mar 12 at 21:40
GurjinderGurjinder
552417
$endgroup$
add a comment |
$begingroup$
Is $log(-x)$, where $x in (0, infty)$ undefined? From solving quadratics I was first told that the discriminant has to be non-negative, since $sqrt{-a}$, where $a in (0, infty)$ is undefined. But this was before learning about Imaginary numbers and the complex plane.
For example, what would $log_3(-9)$ be? Is there a number $a$ such that $3^a=-9$?
So can you evaluate negative logarithms?
real-analysis complex-analysis logarithms
asked Mar 12 at 21:40
GurjinderGurjinder
552417
$endgroup$
2
$begingroup$
You absolutely can, and the extension is due to complex analysis. Assuming the principle branch of $log z$, consider $z=-1=e^{i pi}$ and so $log z = i pi$. In general, $log z = ln|z| + i pi$
$endgroup$
– Ryan Goulden
Mar 12 at 21:43
$begingroup$
in complex plane yes.
$endgroup$
– cand
Mar 12 at 21:44
$begingroup$
It depends whether you allow complex numbers, but even then, the logarithms of negative numbers are not unique, you have to choose a particular branch to make it unique.
$endgroup$
– Peter
Mar 12 at 22:38
add a comment |
$begingroup$
Is $log(-x)$, where $x in (0, infty)$ undefined? From solving quadratics I was first told that the discriminant has to be non-negative, since $sqrt{-a}$, where $a in (0, infty)$ is undefined. But this was before learning about Imaginary numbers and the complex plane.
For example, what would $log_3(-9)$ be? Is there a number $a$ such that $3^a=-9$?
So can you evaluate negative logarithms?
real-analysis complex-analysis logarithms
asked Mar 12 at 21:40
GurjinderGurjinder
552417
$endgroup$
Is $log(-x)$, where $x in (0, infty)$ undefined? From solving quadratics I was first told that the discriminant has to be non-negative, since $sqrt{-a}$, where $a in (0, infty)$ is undefined. But this was before learning about Imaginary numbers and the complex plane.
For example, what would $log_3(-9)$ be? Is there a number $a$ such that $3^a=-9$?
So can you evaluate negative logarithms?
real-analysis complex-analysis logarithms
real-analysis complex-analysis logarithms
asked Mar 12 at 21:40
GurjinderGurjinder
552417
asked Mar 12 at 21:40
GurjinderGurjinder
552417
asked Mar 12 at 21:40
GurjinderGurjinder
552417
asked Mar 12 at 21:40
GurjinderGurjinder
552417
asked Mar 12 at 21:40
GurjinderGurjinder
552417
552417
2
$begingroup$
You absolutely can, and the extension is due to complex analysis. Assuming the principle branch of $log z$, consider $z=-1=e^{i pi}$ and so $log z = i pi$. In general, $log z = ln|z| + i pi$
$endgroup$
– Ryan Goulden
Mar 12 at 21:43
$begingroup$
in complex plane yes.
$endgroup$
– cand
Mar 12 at 21:44
$begingroup$
It depends whether you allow complex numbers, but even then, the logarithms of negative numbers are not unique, you have to choose a particular branch to make it unique.
$endgroup$
– Peter
Mar 12 at 22:38
add a comment |
2
$begingroup$
You absolutely can, and the extension is due to complex analysis. Assuming the principle branch of $log z$, consider $z=-1=e^{i pi}$ and so $log z = i pi$. In general, $log z = ln|z| + i pi$
$endgroup$
– Ryan Goulden
Mar 12 at 21:43
$begingroup$
in complex plane yes.
$endgroup$
– cand
Mar 12 at 21:44
$begingroup$
It depends whether you allow complex numbers, but even then, the logarithms of negative numbers are not unique, you have to choose a particular branch to make it unique.
$endgroup$
– Peter
Mar 12 at 22:38
2
2
$begingroup$
You absolutely can, and the extension is due to complex analysis. Assuming the principle branch of $log z$, consider $z=-1=e^{i pi}$ and so $log z = i pi$. In general, $log z = ln|z| + i pi$
$endgroup$
– Ryan Goulden
Mar 12 at 21:43
$begingroup$
You absolutely can, and the extension is due to complex analysis. Assuming the principle branch of $log z$, consider $z=-1=e^{i pi}$ and so $log z = i pi$. In general, $log z = ln|z| + i pi$
$endgroup$
– Ryan Goulden
Mar 12 at 21:43
$begingroup$
in complex plane yes.
$endgroup$
– cand
Mar 12 at 21:44
$begingroup$
in complex plane yes.
$endgroup$
– cand
Mar 12 at 21:44
$begingroup$
It depends whether you allow complex numbers, but even then, the logarithms of negative numbers are not unique, you have to choose a particular branch to make it unique.
$endgroup$
– Peter
Mar 12 at 22:38
$begingroup$
It depends whether you allow complex numbers, but even then, the logarithms of negative numbers are not unique, you have to choose a particular branch to make it unique.
$endgroup$
– Peter
Mar 12 at 22:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrt{x^2+y^2}$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).
Equipped with this knowledge and the fact that
$log_a z=frac{ln z}{ln a}$:
$$
log_3(-9)=frac{ln9+ipi}{ln3}=2+frac{pi i}{ln3}.
$$
answered Mar 12 at 22:36
useruser
5,54911030
$endgroup$
add a comment |
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$begingroup$
Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrt{x^2+y^2}$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).
Equipped with this knowledge and the fact that
$log_a z=frac{ln z}{ln a}$:
$$
log_3(-9)=frac{ln9+ipi}{ln3}=2+frac{pi i}{ln3}.
$$
answered Mar 12 at 22:36
useruser
5,54911030
$endgroup$
add a comment |
$begingroup$
Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrt{x^2+y^2}$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).
Equipped with this knowledge and the fact that
$log_a z=frac{ln z}{ln a}$:
$$
log_3(-9)=frac{ln9+ipi}{ln3}=2+frac{pi i}{ln3}.
$$
answered Mar 12 at 22:36
useruser
5,54911030
$endgroup$
add a comment |
$begingroup$
Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrt{x^2+y^2}$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).
Equipped with this knowledge and the fact that
$log_a z=frac{ln z}{ln a}$:
$$
log_3(-9)=frac{ln9+ipi}{ln3}=2+frac{pi i}{ln3}.
$$
answered Mar 12 at 22:36
useruser
5,54911030
$endgroup$
Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrt{x^2+y^2}$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).
Equipped with this knowledge and the fact that
$log_a z=frac{ln z}{ln a}$:
$$
log_3(-9)=frac{ln9+ipi}{ln3}=2+frac{pi i}{ln3}.
$$
answered Mar 12 at 22:36
useruser
5,54911030
answered Mar 12 at 22:36
useruser
5,54911030
answered Mar 12 at 22:36
useruser
5,54911030
answered Mar 12 at 22:36
useruser
5,54911030
5,54911030
add a comment |
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$begingroup$
You absolutely can, and the extension is due to complex analysis. Assuming the principle branch of $log z$, consider $z=-1=e^{i pi}$ and so $log z = i pi$. In general, $log z = ln|z| + i pi$
$endgroup$
– Ryan Goulden
Mar 12 at 21:43
$begingroup$
in complex plane yes.
$endgroup$
– cand
Mar 12 at 21:44
$begingroup$
It depends whether you allow complex numbers, but even then, the logarithms of negative numbers are not unique, you have to choose a particular branch to make it unique.
$endgroup$
– Peter
Mar 12 at 22:38