## Complex exponentiation

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### Complex exponentiation

I have recently been reading a few pages similar to this
http://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/

What I don't get is the place where ii is mentioned. True, the exponent turns real with a change of base, but what happened to the intuitive understanding now? We normally obtain positive real numbers by accelerating 1 forward and transforming it and we apply that transformation as many times as the exponent says. If the exponent is negative, we transform it backwards, if it is fractional, we transform it fractionally and if it is imaginary, we transform it sideways and rotate it. That's fine. But how about raising i to the exponent? We get i be rotating 1 to the left by pi/2 radians. Thus for real exponents we have to rotate accordingly and use de Moivre's theorem

(cos x + i sin x)n = cos nx + i sin nx

But what kind of transformation does ii represent?

And is it true that 11 - (i ln 2)/2pi = 2? I got this by taking 1 as e2i.pi and 2 as eln 2 + 2i.pi. Am I allowed to take some non-principal value and transform it like this or will that result in a fallacy like -1 = i2 = i*i = -11/2*-11/2 = (-1*-1)1/2 = 11/2 = 1?

There is much that I have understood about complex exponentiation, though I feel that there is too much that I have not understood.
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Prashantkrishnan
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### Re: Complex exponentiation

First consider a^z = exp(z . ln(a)), thus i^i = exp(i . ln(i)).
Next consider, provided ln(b) = z, then b = exp(z), and thus, because of the identity i = exp(i . pi/2), you'll get ln(i) = i . pi/2.
Thus together: i^i = exp(i . ln(i)) = exp(i . i . pi/2) = exp(-pi/2) = 1/sqrt(exp(pi)),
or just some positive real number with value 0.207 879 576 35...

--- rk
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### Re: Complex exponentiation

Klitzing wrote:First consider a^z = exp(z . ln(a)), thus i^i = exp(i . ln(i)).
Next consider, provided ln(b) = z, then b = exp(z), and thus, because of the identity i = exp(i . pi/2), you'll get ln(i) = i . pi/2.
Thus together: i^i = exp(i . ln(i)) = exp(i . i . pi/2) = exp(-pi/2) = 1/sqrt(exp(pi)),
or just some positive real number with value 0.207 879 576 35...

--- rk

This was very clear in the webpage already... What I was looking for for was the type of transformation. Anyway, I've understood that now . I notice a general rule that if a positive real number is raised to the power of a real number, we get a positive real number and if it is raised to the power of an imaginary number, we get a number on the origin centred unit circle. If a number on the unit circle is raised to the power of a real number, we get a number on the unit circle and if it is raised to an imaginary number, we get a positive real number. But there are still some things I am not understanding . ii can be expressed as ei ln i and ln i has infinitely many values. Thus we find that ii can represent infinitely many real values, with the principal value being 0.207 879 576 35...

Between two values of ii, we may have some value of i2i, i3i, i3i ... and we don't know where these might be and in which order they appear. All this would say much about how the exponentiation of an imaginary number to the power of another imaginary numbers behaves.

The function f(x) = iix must be a continuous function (actually a multivalued function, a misnomer). We can take this as a real valued function with domain R and range R+. This means that in some interval between two consecutive values of ii either there has to be a maximum value of x beyond which the function has no more values in that interval or in every interval there is a value mapping to ii. infinity. The first explanation seems more probable. But both seem too strange. In any case, I would like to see a graph of this function. I suppose it is a 2D graph.

Correct me if I am wrong in any of my assumptions.

By the way,
Prashantkrishnan wrote:11 - (i ln 2)/2pi = 2
is this true or false?
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Prashantkrishnan
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### Re: Complex exponentiation

The graph is not too interesting. Nothing more than a curve, intersecting at y=1, and approaches X at infinity. The graph says Z in place of Y, since I use a 3D plotter.

f(x) = iix
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### Re: Complex exponentiation

ICN5D wrote:The graph is not too interesting. Nothing more than a curve, intersecting at y=1, and approaches X at infinity. The graph says Z in place of Y, since I use a 3D plotter.

f(x) = iix

This seems quite weird... I mean, only the principal values are shown in this graph. Some of the non principal values of ii can be e-5pi/2, e3pi/2 etc. But from the graph, we might find z = e3pi/2 corresponding to a negative value of x. This seems to be one of the least elegant situations in complex exponentiation. This would mean that logie3pi/2 has two values (Actually infinitely many values). Exponential and logarithmic functions are no longer functions in the complex number system.
People may consider as God the beings of finite higher dimensions,
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Prashantkrishnan
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### Re: Complex exponentiation

It is relatively easy to show, that dilation of the ordinary number line represents multiplication by whatever '1' happens to land on.

On the plane, it is equally easy to show, that dialation + rotation of the plane gives a multiplication, where y=ix, and the multiplier is where ever 1 ends up at.

One can for example, prove that AB = BA, and that multiplication is given by R cis(A) * r cis(a) = Rr cis(A+a). cis = cos + i sin

From simple coordinates, you can prove that (C+iS) * (c+is) = Cc + i² Ss + i(Cs + sC), and the coordinate of the point is Cc-Ss + i(Cs-sC), whence i²=-1.

One can by putting e^(ix) expand out the taylor series and set i^2=-1, divides into cos(x)+i sin(x), for all x, where these are done in radians.

From this, one can then find the logrithm of any complex number, up to cycles in 2 pi i, and multiply these, So i^i has many values, since it corresponds to e^(pi(4n+1)i/2) But since e^2pi i = 1, we get i^i = cis(2 pi (n + 1/4)) = cisc(1/4) = i.
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wendy
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### Re: Complex exponentiation

Wendy, I suppose you were struggling here with exp(ix) = cis(x), that somewhere getting some missing i within your exponentiations.

Klitzing wrote:First consider a^z = exp(z . ln(a)), thus i^i = exp(i . ln(i)).
Next consider, provided ln(b) = z, then b = exp(z),

So far this all is just definition, examplification of that definition, resp. mere application of exponentiation. Thus no different values being obtained here so far.

and thus, because of the identity i = exp(i . pi/2), you'll get ln(i) = i . pi/2.

Here we indeed could have more generally: i = exp(i(4n+1)pi/2).

Thus together: i^i = exp(i . ln(i)) = exp(i . i . pi/2) = exp(-pi/2) = 1/sqrt(exp(pi)),
or just some positive real number with value 0.207 879 576 35...

Then we could generalize this to
i^i = exp(i . ln(i)) = exp(i . i (4n+1) pi/2) = exp(-(4n+1) pi/2) = (exp(-pi/2))^(4n+1) = ( 1/sqrt(exp(pi)) )^(4n+1)
i.e. (0.207 879 576 35...)^(4n+1) for any integral n.

--- rk
Klitzing
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### Re: Complex exponentiation

Could do. It's quite hot over here at the moment, (100 F = 35 C, but high humidity), and the big iron does not like the heat. So i am using boxes which it's hard to see and type.

I'm pretty sure that i = cisc(N + 1/4), = exp( 2pi i [N+1/4]). i^i is then exp(-2pi (N+1/4) = 111.31777/535.491^x, according to which cycle you take.
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