what happens to i^(square root of 2)?

I always thought what be in store for us if take irrational number to power of a complex number? we already have natural number, integers, rational numbers, real numbers and complex numbers. Will it create a class of its own?

Note by Shiva Kumar
9 months, 2 weeks ago

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yes sir but when the value of k is increased in polar form of (i)^(square root of 2), we do not get the solutions that superimpose the previous solutions. so we instead get infinite no of solutions. it form a circle with magnitude equal to one and center as origin in Argand Plane after k becomes infinite. so does that mean (i)^(any irrational number) forms a circle with magnitude of value one and center at origin.and get same set of solutions. does this mean (i)^any irrational number has same set of values.

Shiva Kumar - 9 months, 2 weeks ago

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The basic way to define powers of complex numbers is via the formula \[ z^w \; = \; e^{w\log z} \] and so the whole business revolves around the definition of the logarithm of complex numbers. We need \[ \log z \; = \; \ln |z| + i\mathrm{Arg}\,z \] and here is the real problem. There is no way of defining the argument function continuously (let along differentiably) on the whole complex plane.

Since we want the argument (and hence the logarithm) to be differentiable, it has to be defined on an open set (so that we can consider its derivative at all points), and complex functions are considered on open connected domains. The standard way of doing this for the argument is to cut the plane. This means considering the domain formed by the complex numbers with a straight line out from \(0\) to infinity removed. The argument can be defined uniquely (and differentiably) on any such domain, and so can the natural logarithm of the modulus (since we have removed \(0\)).

For example, we could consider the cut plane \(\mathbb{C} \backslash (-\infty,0]\) consisting of the complex numbers except for the nonpositive reals. This means that \(|z| > 0\) and \(-\pi < \mathrm{Arg}\,z < \pi\) for all \(z\) in the cut plane, and we can define \(\log z\) uniquely (and differentiably) now. This choice is called using the principal branch of the argument. In this case we would have \(\ln i = \tfrac12\pi i\), and so \(i^\sqrt{2} = e^{\frac{\pi i}{\sqrt{2}}}\).

A different cut would be to exclude the nonnegative reals \([0,\infty)\), and decide that \(2\pi < \mathrm{Arg}\,z < 4\pi\) for all \(z\) in this cut plane. This would make \(\ln i = \tfrac52\pi\), giving a totally different value for \(i^{\sqrt{2}}\).

This is what @Pi Han Goh means about this function being multivalued. There is no way to have a single function that works everywhere.

The other option is to allow all these values to work at the same time! This means we have to leave the ordinary complex plane, and work with a Riemann surface. Imagine an infinite number of complex planes stacked one on top of another. Cut them all along the positive real axis, and stick the "first quadrant edge" of each sheet to the "fourth quadrant edge" of the sheet below. The resulting helicoidal shape allows you to consider complex numbers with positive moduli and all possible arguments. The number \(i\) would lie in one sheet, the number \(ie^{2\pi i}\) would lie in the sheet above, the number \(ie^{-6\pi i}\) would lie in three sheets below, and so one. You could then define a logarithm on the whole Riemann surface at one go...

Mark Hennings - 9 months, 1 week ago

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Ah crap. I can't answer this question either. My first though is Equidistribution theorem, but I'm not entirely confident.

Summoning the great @Mark Hennings again!

Pi Han Goh - 9 months, 1 week ago

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No, \(i^{\sqrt 2} \) is just a multivalued complex numbers.

Use De-Moivre's theorem.

Let \(x = i^{\sqrt 2} \), then \( \ln x = \sqrt 2 \cdot i = \sqrt 2 ( 0 + 1i ) \).

With \(0 + 1i = \cos \left (\frac\pi 2 + 2\pi k \right) + i \sin \left ( \frac\pi 2 + 2\pi k \right) \), where \(k\) is any integer.

Can you finish it off from here?

Pi Han Goh - 9 months, 2 weeks ago

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