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 Trouble with chapter 5 
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Joined: 15 Apr 2009, 23:40
Posts: 6
Post Trouble with chapter 5
I'm really having trouble understanding chapter 5 in the sense of the inverse of the complex exponential function e. For example, I understand the basic mechanics of logarithms and complex numbers with their polar coordinates, like multiplying them and seeing that you multiply their modulii and add their arguments. However, I don't understand how the argument itself is a logarithm other than it appearing like one in the sum and difference identities in trig. I also don't understand how in w = e^z, the logarithm z = log r + (theta)i


17 Apr 2009, 06:14
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Joined: 07 Jun 2008, 08:21
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Post Re: Trouble with chapter 5
I think I understand what you are having difficulties with - section 5.3 of chapter 5.
So here goes (let me know if this helps or if you have further questions):
If w=e^z, then by definition z=\log w\ \ \ \ \ \ (1).

If we now express w as re^{i\theta} then

z=\log w=\log(re^{i\theta})=\log r+\log(e^{i\theta})

Using (1) means that \log(e^{i\theta})=i\theta, so this finally becomes

z=\log w=\log r+i\theta

So the way I understand what Penrose is saying is that \log w has a real part equal to \log r and an imaginary part equal to \theta the argument of w, and in that sense \theta is a logarithm.

Perhaps to make it more obvious, if we take the case of w=e^{i\theta} (i.e. when r=1) then:

\log w=\log 1+i\theta=i\theta, since \log 1=0.

So \log w=i\theta


17 Apr 2009, 07:22

Joined: 15 Apr 2009, 23:40
Posts: 6
Post Re: Trouble with chapter 5
Oh I see! Thank you so much. I think I was confused by the fact that I wasn't seeing w as e^i^\theta


18 Apr 2009, 05:15

Joined: 15 Apr 2009, 23:40
Posts: 6
Post Re: Trouble with chapter 5
I have another question about chapter 5 and complex powers:
I'm having trouble understanding the solution given on viewtopic.php?f=19&t=100 for exercise 5.9 mainly because I can't seem to tell how we can multiply w^z by e^{z2\pi\mathi} an integer number times and get an allowable w^z.

Is it simply something like this log (w) + 2\pi\imath n = log (wz) if log ( z ) = 2\pi\imath n ? And why is z_2 = p_0 e^{(z-1)\imath t_2} chosen as a second spiral?

I felt like I should remain in this post since you might be monitoring it. Please forgive me for my lack of clarity, this is my first experience with anything in the realm of complex analysis and it all seems a little 'imaginary' to me :-p.

EDIT: I think I realized that z_2 = p_0 e^{(z-1)\imath t_2} was chosen since (z-1)\imath t_2 = z\imath t_2 - \imath t_2 and since t_2 is 2k\pi then the other spiral is just found by subtracting an integral multiple of 2\pi.


18 Apr 2009, 05:33
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Post Re: Trouble with chapter 5
Skynt wrote:
I have another question about chapter 5 and complex powers:
I'm having trouble understanding the solution given on http://roadtoreality.info/viewtopic.php?f=19&t=100 for exercise 5.9 mainly because I can't seem to tell how we can multiply w^z by e^{z2\pi\mathi} an integer number times and get an allowable w^z.

Is it simply something like this log (w) + 2\pi\imath n = log (wz) if log ( z ) = 2\pi\imath n ?

Penrose explains this in section 5.3 of his book, just before figure 5.7 and also after fig 5.7.
It's because the argument of any complex number is ambiguous. You can add 2\pi n,n=0, \pm1,\pm2 etc to \theta (see fig 5.4 in the book) so:
\log w= \log r+i(\theta+2n\pi), n=0, \pm1,\pm2 etc
So w=re^{i(\theta+2\pi n)}=re^{i\theta}.e^{2\pi in},n=0, \pm1,\pm2 etc
So this means that any complex number w=re^{i\theta} can be multiplied by e^{2\pi in}, n=0, \pm1,\pm2 etc and still be a valid value for w.
For this reason e^{w} and \log w are multifunctions (multivalued functions) and strictly speaking represent infinite sets of values.

Skynt wrote:
And why is z_2 = p_0 e^{(z-1)\imath t_2} chosen as a second spiral?

I felt like I should remain in this post since you might be monitoring it. Please forgive me for my lack of clarity, this is my first experience with anything in the realm of complex analysis and it all seems a little 'imaginary' to me :-p.

EDIT: I think I realized that z_2 = p_0 e^{(z-1)\imath t_2} was chosen since (z-1)\imath t_2 = z\imath t_2 - \imath t_2 and since t_2 is 2k\pi then the other spiral is just found by subtracting an integral multiple of 2\pi.

I'm glad you understand it! I have never understood this proof because I didn't understand the same thing as you. I would like to encourage you to submit another proof of this with more explanation if you can.
Vasco


18 Apr 2009, 06:43

Joined: 23 May 2011, 23:51
Posts: 1
Post Re: Trouble with chapter 5
I'm also having trouble when Penrose defines z=ln(r) + i(theta)
Follow me, and see if what I'm doing is wrong.

z=r[cos(theta) + isin(theta)]
z=r[e^(i(theta))] By Euler's Formula
ln(z)=ln[r[e^(i(theta))]]
ln(z)=lnr + i(theta)

Have I done something incorrect? Because I am confident in the obivios fact z =/= ln(z)

And I'm aware this was just previously proved, but I would like some closure as to where my mistake is at.


23 May 2011, 23:58
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Post Re: Trouble with chapter 5
JohnJarboe wrote:
I'm also having trouble when Penrose defines z=ln(r) + i(theta)
Follow me, and see if what I'm doing is wrong.

z=r[cos(theta) + isin(theta)]
z=r[e^(i(theta))] By Euler's Formula
ln(z)=ln[r[e^(i(theta))]]
ln(z)=lnr + i(theta)


Have I done something incorrect? Because I am confident in the obivios fact z =/= ln(z)

And I'm aware this was just previously proved, but I would like some closure as to where my mistake is at.

There is nothing wrong with your maths in the 4 lines in blue above. The maths in red however contradicts what Penrose has written and that's why you are confused.
If you look on page 94 you will see that Penrose defines w as re^{itheta} NOT z.
So z=log w=log{re^itheta}}=log r + itheta
Do you see it now?
Vasco


30 May 2011, 07:41
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