# Tag Info

## New answers tagged complex

8

Your direct method isn't working because neither Complex nor Collect works the way you are expecting. So depending on what you mean by "elegant", the answer to your question may be no. You can jury-rig Collect to work using expr = 2 u + I + 2 + I; Expand[expr/I] /. Complex[x_, y_] -> x + i y I Collect[%, i, Simplify] /. i -> I (* 2 - 2 i - 2 i u *) ...

4

Your plot will trace a series of concentric ellipses, with a point when $r=0$. The eigenvalues have imaginary part zero, and are symmetric about the point $(1,0)$. Start by building a table of output values, here we span $0\leq r \leq 1$ in tenths, and $0 \leq t \leq 2 \pi$ in tenths as well (I am replacing your $\theta$ with $t$ for character simplicity). ...

3

Your question is both basic and broad which means it will probably end up closed unless you can edit it to be more specific. Nevertheless as you are new here is a start: expr := 1 + 3*r*Cos[θ] - I*r*Sin[θ]; Table[ DensityPlot[fn @ expr, {r, 0, 1}, {θ, 0, 2 Pi}], {fn, {Re, Im, Abs, Arg}} ] ~Partition~ 2 // GraphicsGrid Note that capitalization is ...

2

We can use FindAllCrossings[] to find the roots of $\Re\left(\zeta\left(\frac12+i t\right)\right)$ like so: FindAllCrossings[Re[Zeta[1/2 + I t]], {t, 10, 30}, WorkingPrecision -> 20] {14.134725141734693790, 14.517919628262233651, 20.654044969367919453, 21.022039638771554993, 25.010857580145688763, 25.491508214625488621, 29.738510300151580038} ...

6

Using your definition of eq, if you try Solve even without imposing conditions on ω, you obtain the following error: Solve[eq[1, ω] == 0, ω] Solve::inex: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve ...

0

Plot[{Re[(9 Tanh[x]^8 (1 - 2 Tanh[x]^2)^2 (1 - Tanh[x]^2))/(1 + 3 I Sqrt[2/13])^2], Im[(9 Tanh[x]^8 (1 - 2 Tanh[x]^2)^2 (1 - Tanh[x]^2))/(1 + 3 I Sqrt[2/13])^2]}, {x, 0, 10}, PlotRange -> All]

3

Here is a sightly modified version of the utility function I used in an answer to How does Mathematica understand branchcuts of the complex logarithm: ClearAll[plotZ] Options[plotZ] = {ImageSize -> 300, Background -> Lighter[Orange], ContourStyle -> Blue, ExclusionsStyle -> Red, PlotLabel -> Framed[Grid[{{Style["-", Bold, Blue], ...

0

Well, to plot the Abs, use Plot... Plot[Abs[(9 Tanh[x]^8 (1 - 2 Tanh[x]^2)^2 (1 - Tanh[x]^2))/(1 + 3 I Sqrt[2/13])^2], {x, -10, 10}] to plot the real part, replace Abs with Re

4

I can see your surprise, but Mathematica seems to be behaving as you would expect according to the documentation. In particular, it may seem careless to you, but it is clearly spelled out in the documentation that any variable involved in an inequality is implicitly considered a real number. As a reference, see: the Assumptions and Domains page of the ...

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