MHB Evaluating the log of an imaginary number

[burgess]

needed help to solve this math home work? Please help..

What is the value of log(i*pi/2) ?

I know the answer is "i*pi/2", but don't know the procedure to solve it. Please help me.

Thanks a lot in advance.

 

[Prove It]

needed help to solve this math home work? Please help..

What is the value of log(i*pi/2) ?

I know the answer is "i*pi/2", but don't know the procedure to solve it. Please help me.

Thanks a lot in advance.

I tend to avoid the complex logarithm where possible...

$\displaystyle \begin{align*} z &= \log{ \left( \frac{\mathrm{i}\,\pi}{2} \right) } \\ \mathrm{e}^z &= \frac{\mathrm{i}\,\pi}{2} \\ \mathrm{e}^{x + \mathrm{i}\,y} &= \mathrm{i}\,\frac{\pi}{2} \textrm{ where } x, y \in \mathbf{R} \\ \mathrm{e}^x\cos{(y)} + \mathrm{i}\,\mathrm{e}^x\sin{(y)} &= 0 + \mathrm{i}\,\frac{\pi}{2} \\ \mathrm{e}^x\cos{(y)} = 0 \textrm{ and } \mathrm{e}^x\sin{(y)} &= \frac{\pi}{2} \end{align*}$

Solving the first equation we have:

$\displaystyle \begin{align*} \mathrm{e}^x\cos{(y)} &= 0 \\ \cos{(y)} &= 0 \textrm{ as } \mathrm{e}^x > 0\textrm{ for all real }x \\ y &= \frac{(2n + 1)\,\pi}{2} \textrm{ where } n \in \mathbf{Z} \end{align*}$

Substituting into the second equation gives

$\displaystyle \begin{align*} \mathrm{e}^x\sin{(y)} &= \frac{\pi}{2} \\ \mathrm{e}^x \sin{ \left[ \frac{ \left( 2n + 1 \right) \, \pi }{2} \right] } &= \frac{\pi}{2} \\ \mathrm{e}^x \left( -1 \right) ^n &= \frac{\pi}{2} \\ \mathrm{e}^x &= \frac{\pi}{2 \left( -1 \right) ^n } \end{align*}$

This will only be possible where n is even (as a real exponential function is always positive) , so that means we have to let $\displaystyle \begin{align*} n = 2m \textrm{ where } m \in \mathbf{Z} \end{align*}$ and we have

$\displaystyle \begin{align*} \mathrm{e}^x &= \frac{\pi}{2 \left( -1 \right) ^{2m} } \\ \mathrm{e}^x &= \frac{\pi}{2} \\ x &= \ln{ \left( \frac{\pi}{2} \right) } \end{align*}$

Thus $\displaystyle \begin{align*} z = x + \mathrm{i}\,y = \ln{ \left( \frac{\pi}{2} \right) } + \mathrm{i} \, \frac{ \left( 4m + 1 \right) \, \pi }{2} \end{align*}$So $\displaystyle \begin{align*} \log{ \left( \frac{\mathrm{i}\,\pi}{2} \right) } = \ln{ \left( \frac{\pi}{2} \right) } + \mathrm{i}\,\frac{ \left( 4m + 1 \right) \, \pi }{2} \end{align*}$.

 

[Opalg]

needed help to solve this math home work? Please help..

What is the value of log(i*pi/2) ?

I know the answer is "i*pi/2", but don't know the procedure to solve it. Please help me.

Thanks a lot in advance.

If the modulus-argument form of $z$ is $z=re^{i\theta}$ then the principal value of $\log z$ is $\log z = \log r + i\theta$. So for $z = i\pi/2$, what are the modulus and argument of $z$?

[Prove It's answer is correct, but I think that the modulus-argument method is a good deal simpler.]

 

[Prove It]

If the modulus-argument form of $z$ is $z=re^{i\theta}$ then the principal value of $\log z$ is $\log z = \log r + i\theta$. So for $z = i\pi/2$, what are the modulus and argument of $z$?

[Prove It's answer is correct, but I think that the modulus-argument method is a good deal simpler.]

But how do we know they want the principal value?

That was a rhetorical question and illustrates why I prefer to avoid the complex logarithm where possible...

 

[chisigma]

Thefact that every time we speak of the complex logarithm some misunderstandings arise proves to me that in most of the texts it is ill-defined. Of course at this point I am required to provide the'correct definition' and here I report what I have already writtenelsewhere over several years. Consider the Taylor expansion of the function ln z around z = 1 ...$\displaystyle \ln z = 2\ k\ \pi\ i - \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\ (z-1)^{n},\ |z-1|<1\ (1) $

Now the integer constant k in (1) define the brantch of the logarithm function and for k=0 we have the so called 'principal value'. What is very important to say is that once we choose k it remains constant for a complete rotation from 0 to $2\ \pi$. In particular the Taylor expansion in z=-1 can be found simply writing in (1) -z instead of z and is... $\displaystyle \ln z = (2\ k\ + 1)\ \pi\ i - \sum_{n=1}^{\infty}\frac{(z+1)^{n}}{n},\ |z+1|<1\ (2) $

Contrary to what is written in almost all texts the function ln z is analytic at z = -1, as in all other points of the unity circle where k is constant.

Kind regards

$\chi$ $\sigma$

 

[Deveno]

Thefact that every time we speak of the complex logarithm some misunderstandings arise proves to me that in most of the texts it isill-defined. Of course at this point I am required to provide the'correct definition' and here I report what I have already writtenelsewhere over several years. Consider the Taylor expansion of the function ln z around z = 1 ...$\displaystyle \ln z = 2\ k\ \pi\ i - \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\ (z-1)^{n},\ |z-1|<1\ (1) $

I believe this is called the Mercator series, and is a restriction of the function often denoted $\text{Log}(z)$ (the one I believe Opalg is making reference to, implicitly) to the unit disk of radius 1 centered at the complex number 1.

Now the integer constant k in (1) define the brantch of the logarithm function and for k=0 we have the so called 'principal value'. What is very important to say is that once we choose k it remains constant for a complete rotation from 0 to $2\ \pi$. In particular the Taylor expansion in z=-1 can be found simply writing in (1) -z instead of z and is... $\displaystyle \ln z = (2\ k\ + 1)\ \pi\ i - \sum_{n=1}^{\infty}\frac{(z+1)^{n}}{n},\ |z+1|<1\ (2) $

Contrary to what is written in almost all texts the function ln z is analytic at z = -1, as in all other points of the unity circle where k is constant.

This is another function, again restricted to an open disk of radius 1, but this time centered at the complex number -1.

My point is, there is no "the" complex logarithm function. we can only speak of "a" complex logarithm:

$\log:U \to \Bbb C^{\ast}$, where $U$ is a connected open subset of $\Bbb C^{\ast}$.

ANY such function cannot be defined so as to be continuous on a closed curve enclosing the origin.

To highlight the essential problem: any real number $t$ determines a UNIQUE complex number on the unit circle:

$e^{it} = \cos t + i \sin t$

but a complex number on the unit circle does NOT determine a unique $t$, no matter how much you would like it to. It's a funny thing about circles: we can go around them exactly an integral number of times, and wind up going nowhere.

 

[chisigma]

I believe this is called the Mercator series, and is a restriction of the function often denoted $\text{Log}(z)$ (the one I believe Opalg is making reference to, implicitly) to the unit disk of radius 1 centered at the complex number 1.
This is another function, again restricted to an open disk of radius 1, but this time centered at the complex number -1.

My point is, there is no "the" complex logarithm function. we can only speak of "a" complex logarithm:

$\log:U \to \Bbb C^{\ast}$, where $U$ is a connected open subset of $\Bbb C^{\ast}$.

ANY such function cannot be defined so as to be continuous on a closed curve enclosing the origin.

To highlight the essential problem: any real number $t$ determines a UNIQUE complex number on the unit circle:

$e^{it} = \cos t + i \sin t$

but a complex number on the unit circle does NOT determine a unique $t$, no matter how much you would like it to. It's a funny thing about circles: we can go around them exactly an integral number of times, and wind up going nowhere.

It's nice to see that sometimes turn out to be the most difficult concepts easy to understand if faced with basic tools. We refer to the figure below ...

View attachment 2677

The black circle, centered at $z_{0}=1$, defines the region of convergence of the Taylor expansion ...

$\displaystyle \ln z = 2\ k\ \pi\ i - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\ (z-1)^{n}\ (1)$

And 'well known that (1) allows the calculation of the function ln z and all its derivatives at any point inside the black circle... for example the point $ z_ {1} = \frac {\sqrt {2}} {2} \ (1 + i) $. This is fortunate because we can now expand the function ln z in $ z = z_ {1} $ and this new expansion will converge inside the red circle and so we have extended the feature outside of the black circle. A further rotation of $ \frac {\pi} {4} $ allows us to expand the function z = i and two further expansions of $ \frac {\pi} {4} $ to expand at z = -1 and this latest expansion converges inside the circle turquoise. Needless to say, proceeding in this way extends the domain of the function ln z to the whole complex plane with the exception of z = 0, which is the only singularity ...

If anyone has any criticisms to make this method [I certainly hope not ...], please keep in mind that the same way You get to prove the existence of the Riemann Zeta Function on the whole complex field with the only exception z = 1 ... and I am certain that he will not deprive many valuable members of the forum of the opportunity to earn a million dollars ...

Kind regards

$\chi$ $\sigma$

 

[Deveno]

It's nice to see that sometimes turn out to be the most difficult concepts easy to understand if faced with basic tools. We refer to the figure below ...

View attachment 2677

The black circle, centered at $z_{0}=1$, defines the region of convergence of the Taylor expansion ...

$\displaystyle \ln z = 2\ k\ \pi\ i - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\ (z-1)^{n}\ (1)$

And 'well known that (1) allows the calculation of the function ln z and all its derivatives at any point inside the black circle... for example the point $ z_ {1} = \frac {\sqrt {2}} {2} \ (1 + i) $. This is fortunate because we can now expand the function ln z in $ z = z_ {1} $ and this new expansion will converge inside the red circle and so we have extended the feature outside of the black circle. A further rotation of $ \frac {\pi} {4} $ allows us to expand the function z = i and two further expansions of $ \frac {\pi} {4} $ to expand at z = -1 and this latest expansion converges inside the circle turquoise. Needless to say, proceeding in this way extends the domain of the function ln z to the whole complex plane with the exception of z = 0, which is the only singularity ...

If anyone has any criticisms to make this method [I certainly hope not ...], please keep in mind that the same way You get to prove the existence of the Riemann Zeta Function on the whole complex field with the sole exception z = 1 ... and I am certain that he will not deprive many valuable members of the forum of the opportunity to earn a million dollars ...

Kind regards

$\chi$ $\sigma$

Indeed this process is called "analytic continuation" and is, as you say, "well-known". However, one finds, if one attempts to "close the bouquet" around the origin, that the values on the original black circle do not agree with the continuation from the "blue circle".

The Riemann surface for the $\log$ function makes it clear what the problem is: we are using the wrong domain; the complex exponential $e^{it}$ should not map to a CIRCLE, but a HELIX, and it is on the "joining" of these nested helices, rather than nested disks, that we ought to define the logarithm.

This, then, totally resolves the ambiguity inherent in:

$\log(re^{i\theta}) = \ln r + i(\theta (+2k\pi))$

and also shows why your individual open disks work (for any given $k$), they are the relevant projections onto the complex punctured plane from the Riemann surface (the projection identifies all successive windings around the origin).

Of course, if we are just interested in "a" solution involving inverting a (complex) exponential function, rather than "all" solutions, we are free to use whatever domain is convenient, presumably a (sufficiently small) open disk around the point of interest, say (for example):

$i\dfrac{\pi}{2}$