How do I simplify i to the i? (1 Viewer)

[tiggerfamilytre]

My teacher couldn't do this, and I don't know where to begin. How do I simplify i to the i?

 

[Slidey]

http://mathforum.org/library/drmath/view/53907.html

The critical line of that link is:

i^i = e^[0 + i^2 pi/2+ i^2 2kpi] = e^[- pi/2 - 2kpi]

That is: i^i has many values

 

[Trev]

wtf?!
ohh this thing, isnt this eulers theorem or something? i asked my math teacher bout this he didnt kno wat it was and he did 4u for hsc lol i just remember it got him thinking about pi and how pi is rational or something (i forget)

 

[Trev]

haha that site "I have been playing around with this equation and have
found some disturbing things." mmmmmyes disturbing mathematics

 

[shannonm]

e^(ix) = cosx + isinx
when x = pi/2,
e^(i.pi/2) = i (cos term = 0, sin term = i(1))
so i^i = e^(-pi/2)

 

[shannonm]

imo this is a very very trivial result, you didn't specify which teacher you did ask, so for the sake of you and other students in your maths class, i hope you asked your business teacher or something

 

[Slidey]

e^(ix) = cosx + isinx
when x = pi/2,
e^(i.pi/2) = i (cos term = 0, sin term = i(1))
so i^i = e^(-pi/2)

Pick any two values of i^i and you're a winner!

 

[Trev]

imo this is a very very trivial result, you didn't specify which teacher you did ask, so for the sake of you and other students in your maths class, i hope you asked your business teacher or something

eh? u tripper i sed i asked my math teacher (if he could prove it)

 

[Slidey]

Trev, the point is a 4u maths teacher has studied mathematics at university. Something is fundamentally wrong if they do not know Euler's wonder-formula e^(i.@)=cis@.

 

[shannonm]

I was reffering to tiggerfamilytre's post not yours.

trev if your maths teacher doesnt know eulers formula he shouldnt be teaching (or was he reffering to the complex number 'i' when he said he didnt know what it was?)





edit-slidey beat me to it but the point remains

 

[Trev]

he doesnt kno eulers thorem i think, i kno he cant prove it coz i remember that much

 

[Captain pi]

he doesnt kno eulers thorem i think, i kno he cant prove it coz i remember that much

Di Bona? I think he would know.

 

[Trev]

i asked him, aaages ago if he could prove it and he didnt kno how to, he probly knows tho. this arv he was goin on about how rosie is the iron chef and he is the guest chef or something, coz he never beats rosie lol

 

[Idyll]

e^(ix) = cosx + isinx
when x = pi/2,
e^(i.pi/2) = i (cos term = 0, sin term = i(1))
so i^i = e^(-pi/2)

with a few quick changes.....


e^(ix) = cosx + isinx
when x = 5pi/2,
e^(i.5pi/2) = i (cos term = 0, sin term = i(1))
so i^i = e^(-5pi/2)


see i^i is actually e^(-5pi/2)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

 

[martin]

All right,
I don't think this result is especially trivial. Be nice to your maths teachers, when you haven't done complex analysis for years its easy to forget little tricks like these.
Basically to understand this question you just have to know Euler's formula
e^ix = cosx + isinx
so i = e^(i*(4k+1)Pi/2)
because we need cosx=0 and sinx=1 which is true at Pi/2,5Pi/2,...

Now you need to accept that the complex exponential function follows the same rules as the real exponential function ie (e^x)^y = e^(xy). There should be a proof of this somewhere on the net.

then
i^i
= e^(i^2*(4k+1)Pi/2)
= e^(-(4k+1)Pi/2)
= e^(-Pi/2-2kPi)

So the result is right but there is some subtlety involved. To see that you need to prove a result about the complex exponential consider the same proof using cisx notation

i^i
= (cos(4k+1)Pi/2 + isin(4k+1)Pi/2)^i
and it definitely isn't obvious how to proceed. The e^z notation is used to make clear the properties of the complex exponential function.

 

[CrashOveride]

Trev, the point is a 4u maths teacher has studied mathematics at university. QUOTE]

This isn't always the case .___.

 

[Will Hunting]

Is it peculiar that I've spent a term plumbing the depths of Complex Numbers and have never heard of a Euler's theorum. I've been using a couple of Patel's publications as references and the said theorum has failed to present itself as yet. Please shed some light on this.

 

[Slidey]

It was removed from the syllabus ages ago, I think.

Basically, r.e^(i.@)=r.cis@
Very useful - you tend to use the exponential form in complex analysis - which is analysis (calculus) of complex numbers.

Somebody who's actually done complex analysis (martin?) might be able to shed some more light. It'll be 3 or so years before I can take it.

 

[Arkad]

Basically, r.e^(i.@)=r.cis@

Basically thats what you get taught in first year uni for complex no. questions.

 

[martin]

To be perfectly clear, none of the following is relevant in the NSW 4U maths course.

In first year uni we only did a week or two of complex numbers but Euler's formula was used a lot. I wondered why we didn't learn it in high school. I guess its difficult to give an honest proof and its very difficult to just accept it.

The standard proof uses Taylor series (infinite series that add together to give sin, cos or exponential functions) but there is another proof that uses differentiation and integration. All you have to accept is that you can differentiate and integrate complex functions as if i is a constant.

Here it is (ripped shamelessly from wikipedia)

Define the complex number z such that

z=cos x + i*sin x

Differentiating z with respect to x:

dz/dx=-sin x + i*cos x

Using the fact that i^2 = -1:

dz/dx=i^2*sin x + i*cos x=i*(cos x + i*sin x)=z*i

Separating variables and integrating both sides:

integral(1/z)dz=integral(i)dx
ln z=xi + C

where C is the constant of integration. To finish the proof we have to argue that it is zero. This is easily done by substituting x = 0.

lnz = C

But z is just equal to:

z = cosx + i*sinx = cos0 + i*sin0 = 1

thus

ln1 = C
C = 0

So now we just exponentiate

lnz = xi
e^lnz = e^xi
z = e^xi
e^xi = cosx + isinx