Euler's Formula (Complex Analysis) Problem

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PrimeLime
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e=1

For what values of § does this hold?

I'm not actually looking for the solution to the problem - I know that you just write cos§ + isin§ = 1 and then equate Re and Im, giving you §=2k(pi) for any integer k.
But the first thing I did when I saw this problem was to take the natural logarithm of both sides, giving i§=0, so §=0. Where did all the other solutions go? Is there something insufficient about taking the natural log of both sides, or maybe ln(1) has more than one value in complex numbers?
Basically, why does the method of taking logarithms not work?
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StrangeBanana
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Yeah, the complex logarithm (i.e. the inverse of the complex exponential function) is multi-valued.
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FireGarden
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As StrangeBanana says, the logarithm does not behave so nicely on C.

For an example;  \log(1+i) = \log(\sqrt{2}e^{i\frac{\pi}{4}})  = \log(\sqrt{2})+i(\dfrac{\pi}{4} + 2n\pi)\quad \forall n\in\mathbb{Z}

So the multivalued-ness is just a result of the fact that an argument for a complex number is not unique. Though there is the notion of a principal argument, \arg(z) \in (-\pi, \pi]. Use this argument above, and you get what's called the principal logarithm. It's important to remember that the principle logarithm is still only one possible value, but the most sensible to pick in some sense. Forgetting this will mean solutions will hide from you.
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PrimeLime
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(Original post by FireGarden)
As StrangeBanana says, the logarithm does not behave so nicely on C.

For an example;  \log(1+i) = \log(\sqrt{2}e^{i\frac{\pi}{4}})  = \log(\sqrt{2})+i(\dfrac{\pi}{4} + 2n\pi)\quad \forall n\in\mathbb{Z}

So the multivalued-ness is just a result of the fact that an argument for a complex number is not unique. Though there is the notion of a principal argument, \arg(z) \in (-\pi, \pi]. Use this argument above, and you get what's called the principal logarithm. It's important to remember that the principle logarithm is still only one possible value, but the most sensible to pick in some sense. Forgetting this will mean solutions will hide from you.
So are you saying that ln(1) is multi-valued as long as nonreal values are allowed?
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FireGarden
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(Original post by PrimeLime)
So are you saying that ln(1) is multi-valued as long as nonreal values are allowed?
Exactly. You can do it.. \log(1) = \log(1 e^{i\cdot 0}) = \underbrace{\log(1)}_\text{Real log} + 2ni\pi = 2ni\pi,\quad n\in\mathbb{Z} .

If only real values are allowed, you get just log(1) (=0) uniquely.


Edit: Perhaps it would be good to try and prove that \text{Log}(z) = \log(|z|) + i(\arg(z) + 2n\pi), \forall n\in\mathbb{Z} where Log = complex log and log = real log.
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PrimeLime
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(Original post by FireGarden)
Exactly. You can do it.. \log(1) = \log(1 e^{i\cdot 0}) = \underbrace{\log(1)}_\text{Real log} + 2ni\pi = 2ni\pi,\quad n\in\mathbb{Z} .

If only real values are allowed, you get just log(1) (=0) uniquely.


Edit: Perhaps it would be good to try and prove that \text{Log}(z) = \log(|z|) + i(\arg(z) + 2n\pi), \forall n\in\mathbb{Z} where Log = complex log and log = real log.
Thanks for your help.
However, I'm still struggling to understand why ln(e^i0)=2ni(pi)? Surely e^i0 would just be evaluated as 1 first?
Oh and if you don't mind, how do you manage to type maths symbols like that?
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FireGarden
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(Original post by PrimeLime)
However, I'm still struggling to understand why ln(e^i0)=2ni(pi)? Surely e^i0 would just be evaluated as 1 first?
An issue with the complex numbers (but also with other number systems) is that we can't write numbers uniquely.

In \mathbb{C}, we can write 1=e^{i0}=e^{i2\pi}=e^{i4\pi}= \ldots = e^{i2n\pi}=\ldots They all represent the same number, which is obvious if you look on an Argand diagram. So when you want to evaluate Log, well.. which of those presentations of the number do you choose? 1 is the "principal" choice as it's the number with a principal argument, but this is really not much more than a fudge to make a certain choice the standard one. Putting all presentations on even footing, we get different results for each presentation, and we can't choose any one over another - so it's just multivalued (and so not really a function.. but people work around this in various ways. One such way is to use the principal value!).

I said similar things happen in other number systems. I'm sure you're aware 1=0.9999... . More exotic-looking things happen in finite fields, like in \mathbb{F}_5 (the field on 5 elements), you'll get 3=-2.

Oh and if you don't mind, how do you manage to type maths symbols like that?
It's called LaTeX (lay-tek. The 'X' is supposed to be a greek chi). TSR has tags , [te x] and then the end tag [/te x] (without the spaces of course) which then interpret the code inside and re-typesets it as mathematics. If you look at people's posts like you can when you quote somebody, the code will appear.

The usual ways of writing maths still works. x^2+y^2=r^2 becomes x^2+y^2=r^2. But some things are different. TSR has a guide to latex here: http://www.thestudentroom.co.uk/wiki/LaTex
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poorform
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As others have stated taking logarithms of a complex number is not as easy in \displaystyle \mathbb{C}. The formula for working out \displaystyle \log(Z) where \displaystyle z is a complex number is derived like this.

\displaystyle \log(z) is any \displaystyle w \in \mathbb{C} such that \displaystyle e^w=z. Since \displaystyle w \in \mathbb{C} we can write \displaystyle w=x+iy for some real numbers \displaystyle x,y

so \displaystyle e^w=e^{x+iy}=e^x+e^{iy}=e^x(\cos  (\theta)+i\sin(\theta)).

Also we can write \displaystyle z=e^w=|z|(\cos(\phi)+i\sin(\phi)  ). (Think of any number on an argand diagram and it should be clear why this is true.)

And therefore \displaystyle e^x(\cos(\theta)+i\sin(\theta))=  |z|(\cos(\phi)+i\sin(\phi)) so \displaystyle e^x=|z| \implies x=\log |z| and \displaystyle \phi=\theta +2k\pi for some integer \displaystyle k (this is where we can get infinitely many values for the log because we can just keep adding 2pi and it is equivalent.)

Hence \displaystyle w=\log(z)=\log |z|+i(\theta +2k\pi) and we get the result that \displaystyle \log(z)=\log |z|+i(\theta +2k\pi)=\log |z|+iArg (z) which has been stated above I believe.

Also it is worth noting that \displaystyle Log(z) with a capital L takes the principal argument for \displaystyle \theta .

Hopefully this makes some sense to you. I only just learned this recently so typing it up helps me understand as well .

Something interesting is that you can actually evaluate logs of negative numbers using this formula.

For example \displaystyle Log(-1)=\log |-1|+i(\pi)=i\pi

(Original post by PrimeLime)
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PrimeLime
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(Original post by FireGarden)
An issue with the complex numbers (but also with other number systems) is that we can't write numbers uniquely.

In \mathbb{C}, we can write 1=e^{i0}=e^{i2\pi}=e^{i4\pi}= \ldots = e^{i2n\pi}=\ldots They all represent the same number, which is obvious if you look on an Argand diagram. So when you want to evaluate Log, well.. which of those presentations of the number do you choose? 1 is the "principal" choice as it's the number with a principal argument, but this is really not much more than a fudge to make a certain choice the standard one. Putting all presentations on even footing, we get different results for each presentation, and we can't choose any one over another - so it's just multivalued (and so not really a function.. but people work around this in various ways. One such way is to use the principal value!).

I said similar things happen in other number systems. I'm sure you're aware 1=0.9999... . More exotic-looking things happen in finite fields, like in \mathbb{F}_5 (the field on 5 elements), you'll get 3=-2.



It's called LaTeX (lay-tek. The 'X' is supposed to be a greek chi). TSR has tags , [te x] and then the end tag [/te x] (without the spaces of course) which then interpret the code inside and re-typesets it as mathematics. If you look at people's posts like you can when you quote somebody, the code will appear.

The usual ways of writing maths still works. x^2+y^2=r^2 becomes x^2+y^2=r^2. But some things are different. TSR has a guide to latex here: http://www.thestudentroom.co.uk/wiki/LaTex
Ok, thanks for your help .
You seem to know a lot of maths; are you an undergraduate now and if so, are you studying maths?
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PrimeLime
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(Original post by poorform)
As others have stated taking logarithms of a complex number is not as easy in \displaystyle \mathbb{C}. The formula for working out \displaystyle \log(Z) where \displaystyle z is a complex number is derived like this.

\displaystyle \log(z) is any \displaystyle w \in \mathbb{C} such that \displaystyle e^w=z. Since \displaystyle w \in \mathbb{C} we can write \displaystyle w=x+iy for some real numbers \displaystyle x,y

so \displaystyle e^w=e^{x+iy}=e^x+e^{iy}=e^x(\cos  (\theta)+i\sin(\theta)).

Also we can write \displaystyle z=e^w=|z|(\cos(\phi)+i\sin(\phi)  ). (Think of any number on an argand diagram and it should be clear why this is true.)

And therefore \displaystyle e^x(\cos(\theta)+i\sin(\theta))=  |z|(\cos(\phi)+i\sin(\phi)) so \displaystyle e^x=|z| \implies x=\log |z| and \displaystyle \phi=\theta +2k\pi for some integer \displaystyle k (this is where we can get infinitely many values for the log because we can just keep adding 2pi and it is equivalent.)

Hence \displaystyle w=\log(z)=\log |z|+i(\theta +2k\pi) and we get the result that \displaystyle \log(z)=\log |z|+i(\theta +2k\pi)=\log |z|+iArg (z) which has been stated above I believe.

Also it is worth noting that \displaystyle Log(z) with a capital L takes the principal argument for \displaystyle \theta .

Hopefully this makes some sense to you. I only just learned this recently so typing it up helps me understand as well .

Something interesting is that you can actually evaluate logs of negative numbers using this formula.

For example \displaystyle Log(-1)=\log |-1|+i(\pi)=i\pi
This was very helpful, thanks to both of you .
I'm definitely going to take complex analysis as one of my modules at uni now .
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FireGarden
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(Original post by PrimeLime)
Ok, thanks for your help .
You are very welcome.

You seem to know a lot of maths; are you an undergraduate now and if so, are you studying maths?
I'm a postgraduate masters student in mathematics. It doesn't feel like I know that much! Most of what people often ask here I'm far too rusty with to be much help (ugh, trig identites.. not the most memorable things), or sometimes its stuff I never studied much because I find them really dull (for example, I'd be useless at BMO because Euclidean geometry, number theory and combinatorics etc. are precisely the things I never cared/learned about..). I'm mostly interested in modern algebra, with bits of analysis and topology thrown in there.
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poorform
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(Original post by PrimeLime)
This was very helpful, thanks to both of you .
I'm definitely going to take complex analysis as one of my modules at uni now .
No problem.
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