Log z = ln r + iQ (r > 0, - p < Q < p)

Is called the

*principal branch*.
So ordinarily this would also suffice to answer the problem but not so fast. The problem is the branch cut designating a ray: Q = 2 p, yields a multiple-valued function, which we prefer not to have for the Principal branch. To see this, consider multiple rotations, yielding Q = 2 p, 4 p, 6 p, etc. so that each of these yields an additional value. Therefore, to obtain a single-valued Principal branch we need to restrict Q, i.e. so that the answer is instead:

Log z = ln r + iQ (r > 0, 0 £ Q < 2p)

Log z = ln r + iQ (r > 0, 0 £ Q < 2p)

2) The sort of diagram - as from a previous blog - is shown in Fig. 2. This was originally for a radius r of 5, with incremental values along the vertical axis. But disregarding those, it can also serve to satisfy the answer for problem 2 of the problem set. Thus, the value a = 1 can be considered to be at the radius indicated by the dark orange circle. The value a = 0.5 is just before it, and the value a = 2 would be the last green (inner) circle near the periphery.

The rays to satisfy the conditon: q = a

in the diagram, these can be for any of the angles shown, for example, q = 30 degrees or p/6, and maybe use p/3 (60 degrees) and p/2 (90 degrees) too.

3) We are given: Q1(a1,t1) = 6 p/5 and Q2(a2,t2) = 7 p/5

Let z1 be the configuration at (a1, t1) and z2 be the configuration at (a2, t2). Then, to express the proper motion undergone by the minor spot we may write:

D Q1Q2 = ln z2 – ln z1

where ln z1 = ln r1 + i6 p/5 and ln z2 = ln r2 + i7 p/5

where ln z1 = ln r1 + i6 p/5 and ln z2 = ln r2 + i7 p/5

D Q1Q2 =

[ln r2 + i7 p/5 ] – [ln r1 + i6 p/5] =

ln r2 – ln r1 + i[arg z2 – arg z1] = ln r2 - ln r1 + i[7 p/5 - 6 p/5 ]

ln r2 – ln r1 + i[arg z2 – arg z1] = ln r2 - ln r1 + i[7 p/5 - 6 p/5 ]

= ln r2 - ln r1 + i(p/5)

The task of further simplification (hint: using the form exp iQ) is left to industrious math mavens!

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