"Principal Branch Square Root of z in Domain C-{0}

In summary, the question is whether the principal branch square root of z has a Laurent series expansion in the domain C-{0}. It is stated that there is a Laurent series expansion for z^(1/2) in C-{0}, but it is not defined at 0 due to the derivative being 1/z^[(2n+1)/2]. The conversation then discusses the details of the principal branch and its analytic nature, and concludes that the principal branch is the principal root with an argument between -pi and pi. The answer is no, as multifunctions do not have Laurent series about their branch-points.
  • #1
ryou00730
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Homework Statement


Does the principal branch square root of z have a Laurent series expansion in the domain C-{0}?

The Attempt at a Solution



Well I'm not really sure what a principal branch is? I believe that there is a Laurent series expansion for z^(1/2) in C-{0} because originally our only problem is that when we take derivative of z^1/2 we get 1/z^[(2n+1)/2] and this is not defined at 0, but is everywhere else... so I think the answer is yes to this, but again I'm unsure of the details of principal branch?
 
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  • #2
No. The square root is a multifunction and these functions do not have Laurent series about their branch-points because they're not fully analytic in a punctured disc surrounding the branch-point, specifically not so along their branch-cuts
 
  • #3
But does taking the principal branch of square root z not deal with that? Does the principal branch mean we only take the principal roots of z?
 
  • #4
The principal branch is analytic except along it's branch-cut which extends out from the origin so that we do not have an analytic domain in a punctured disk surrounding the origin and yes, the principal branch is the principal root with arg between -pi and pi
 

1. What is the Principal Branch Square Root of z in Domain C-{0}?

The Principal Branch Square Root of z in Domain C-{0} is a mathematical concept that refers to the unique square root of a complex number z, where z is a member of the complex plane excluding the origin (0).

2. How is the Principal Branch Square Root of z in Domain C-{0} calculated?

The Principal Branch Square Root of z in Domain C-{0} is calculated using the formula: √z = √|z| * (cos(θ/2) + i*sin(θ/2)), where |z| is the modulus (or absolute value) of z and θ is the argument (or angle) of z.

3. Why is the Principal Branch Square Root of z in Domain C-{0} important?

The Principal Branch Square Root of z in Domain C-{0} is important in complex analysis and other areas of mathematics, as it allows for the calculation of square roots of complex numbers in a consistent and well-defined manner.

4. Can the Principal Branch Square Root of z in Domain C-{0} be negative?

No, the Principal Branch Square Root of z in Domain C-{0} cannot be negative. This is because the square root function is defined as the positive value that, when squared, gives the original number. Therefore, the Principal Branch Square Root of z in Domain C-{0} is always a positive number.

5. How does the Principal Branch Square Root of z in Domain C-{0} relate to the concept of complex conjugates?

The Principal Branch Square Root of z in Domain C-{0} is related to complex conjugates in that if z = a + bi, where a and b are real numbers, then the complex conjugate of z is a - bi. This means that the Principal Branch Square Root of z in Domain C-{0} is also the square root of the complex conjugate of z.

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