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8

The following is not particularly fast and could be more accurate but does make progress toward the goals set in the Question. To begin, consider the analytical properties of f, as defined in the Question. By observation, it has branch points at {I Sqrt[6/5] ξ, I Sqrt[2/5] ξ} and their conjugates. Poles are obtained by p /. ...


3

Let, $$ f=\frac{z-e^{i \pi/2 n}}{z^{2 n}+1} $$ Then, in Mathematica code, the following seems to work, Limit[f, z -> Exp[(I Pi)/(2 n)], Assumptions -> n >= 1/2] which gives the result, $$ -\frac{e^{i \pi/2 n}}{2n} $$ The problem isn't that Mathematica is handling the $0/0$ limit incorrectly. It's that Mathematica, without additional information, ...


4

Generate complex points cc on the unit circle, then map them with your function $f(z)$. Plot the unit circle and its image together, while manipulating the four parameters. With[{cc = CirclePoints[1000.].{1, I}}, Manipulate[ ListLinePlot[ {ReIm[cc], ReIm[cc*(cc - a.{1,I})/((cc - b.{1,I})*(cc - c.{1,I})*(cc - d.{1,I}))] ...


1

If your aim is numerical and graphical I would not "worry" about the cosmetics of the form. In the following I have made k=1 (no values for $\delta$'s given). to avoid numerical problems just due to extreme scales (v precision). I have tried to "correct" the unbalanced parentheses referred to by Bob Hanlon. I may have made an error. If so, I apologize: ...


0

You could also use gc=Conjugate[g[θ]]/.Conjugate[cos_Cos]->cos and h=g*gc and use h as the integrand or plotted function. One should avoid calling Simplify inside Plot or Integrate because this will be much slower than necessary. Also, your integrand contains a factor 10^-30. So it is practically zero and nothing is plotted and the integral is zero.


2

You can use Simplify with the assumption that θ is an element of the Reals. f = Conjugate[Cos[θ]] Simplify[f, θ ∈ Reals]


1

The comments provide very nice answers. Perhaps you wish the 2 x 2 analogs of 1 and I: r = IdentityMatrix[2]; i = {{0, -1}, {1, 0}}; fun[z_] := {r, i}.ReIm[z] Note i.i yields {{-1,0},{0,-1}}= -IdentityMatrix[2]. Look at the comments as they show very nice ways of achieving this.


1

Many thanks to @J.M. and @Artes I found out that the iterator 0.1 in the Table function was making this artifact, which is also version dependent. @J. M.'s solution worked perfectly: Table[N[alpha[y1, d1]], {d1, 1/10, 5, 1/10}, {y1, 0, d1, 1/10}]


1

The trouble with many methods is that they only work on integer inputs. Trying Alexei's answer with approximate numbers {1.0, 1.0 + 2 I, 3.0 - 5 I, 7, 9.0 + I} /. x_ /; Head[x] == Complex -> Sqrt[Re[x]^2 + Im[x]^2]*Exp[I ArcTan[Re[x], Im[x]]] (* {1., 1. + 2. I, 3. - 5. I, 7, 9. + 1. I} *) just spits back out the original answer. Also, a simpler ...


1

Try this: {1, 1 + 2 I, 3 - 5 I, 7, 9 + I} /.x_ /; Head[x] == Complex -> Sqrt[Re[x]^2 + Im[x]^2]*Exp[ArcTan[Re[x], Im[x]]] yielding (* {1, Sqrt[5] E^ArcTan[2], Sqrt[34] E^-ArcTan[5/3], 7, Sqrt[82] E^ArcTan[1/9]} *) Edit: to address your question If you want to have the argument shown as fractions of Pi sa for the angles of 45 grad or 60 ...


0

Since the arguments are rational: v = {1/10, -1 - 2 I, 3 - 5/3 I, 7, 9/10 + I}; Abs[v] Exp[I Arg[v]] {1/10, Sqrt[5] E^(I (-π + ArcTan[2])), 1/3 Sqrt[106] E^(-I ArcTan[5/9]), 7, 1/10 Sqrt[181] E^(I ArcTan[10/9])}



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