Questions tagged [complex]

Questions about using complex numbers in Mathematica. This includes basic arithmetic, functions of complex numbers, plotting complex functions, and dealing with branch cuts.

1337 questions
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Why are Exp[3] and 2 treated differently within Complex?

Why doesn't the last command below split the complex number into its real and imaginary parts? Complex[2,2] (* 2 + 2 I *) Complex[N[Exp[3]], N[Exp[3]]] (* 20.0855 + 20.0855 I *) Complex[Exp[3],Exp[3]) (* Complex[E^3,E^3] *)
Soldalma
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What is the value Re[Sqrt[1+I*2*x]]?

When I try to evaluate Re[Sqrt[z]], for some values of Mathematica fails to evaluate it. For example, Re[Sqrt[2 + I*x]]` Re[Sqrt[2 + I x]] How can I get the real number in Sqrt[complex number]?
user132682
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Modify real part and leaves imaginary part unchanged

How can I flip the sign of the real part but not affect the imaginary part of a complex number: a+bi => -a + bi Example list: list = {{-0.282095 + 0.282095 I, -0.27254 + 0.291336 I, -0.262018 + 0.300835 I, -0.250437 + 0.310542…
leon365
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ComplexExpand absolute squared

ComplexExpand[Abs[a + b I]] Gives $\sqrt{a^2 + b^2 }$ ComplexExpand[Abs[a + b I]^2] On the other hand gives Abs[a + I b]^2 How can I let it evaluate to $a^2 + b^2$ instead?
sjdh
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Visualizing branch cut and Riemann surface for a square root

I have the following complex function: $$g (z) = (1 - a^2/z) (1 - 1 /z),$$ where $0 < a < 1$. Calculations show that $\sqrt{g(z)}$ has a branch cut along $a^2 \to 1$. Is there a way to visualize the Riemann surface of this function and the…
user95917
6
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1 answer

How to make the imaginary part of a +0. I zero globally?

Values like a +0. I are really annoying. Answers from How to reduce expressions with complex coefficients in the form of a+0.*I? Is there a way to globally set when to treat a very small number as zero? give me some clue, but I find $Post = Chop[#,…
novice
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presenting a real number as real instead of imaginary

I have an equation which results in an answer of the form $\frac{i a}{\sqrt{c-d}}$ is there any way to get Mathematica to present it in it's real form? like $\frac{a}{\sqrt{d-c}}$ I know that I can simply take Re[f[x]] to get the real part, but I…
drjrm3
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Complex Solutions not in a+bi form

Solve[x^2 + x + 1 == 0, x] gives the solution: {{x -> -(-1)^(1/3)}, {x -> (-1)^(2/3)}} I know these are complex solutions but I'd like to have them in a+bi form. If I do N@Solve...I get the right form but as you can imagine I get numeric…
user2686410
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3
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Express a complex function $f(z)$ as $u(x,y)+iv(x,y)$

How can I write a complex function $f(z)$ in the form $u(x,y)+iv(x,y)$ using Mathematica? Re and Im do not work, because they do it for complex numbers, not functions. For example, if I try Re[2 I (z) + 6 Conjugate[(z)] + Sin[(z)]] Mathematica…
Thiago
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Absolute value of a complex number

I want to calculate |xx|^2. Why can't Mathematica just calculate it? What am I supposed to do here? I just want it to actually calculate it, so that there are no imaginary numbers left. Is that so hard? xx = (2 I E^(-I a k) k k1)/(2 I k k1 Cosh[a…
Σ balls
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How to get rid of the absolute value in ComplexPlot3D[]?

I'm starting to study complex analysis,so I'm trying out Mathematica as tool for visualizing complex functions. I'm trying to graph $f(z)=z^2 + 1$ with the code: ComplexPlot3D[z^2 + 1 , {z, 2}] and what I get is this: Something is wrong, because I…
3
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2 answers

Find the solution to an equation and show graphically that these lie on a circle in the complex number plane

I need to find the solution to the equation in the picture, and show graphically that these lie on a circle in the complex number plane. How would one go about this?
3
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2 answers

Plotting transformations of loci in a complex plane

I found the solution to plot loci of complex numbers here, e.g. a locus of points of $z$ given $|z|=1$ is a circle centered at the origin with a radius 1. But I am struggling to plot transformations, e.g. $T: w=2z$ would be an enlargement of the…
Mihail
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Incorrect answer of FunctionAnalytic

A new command FunctionAnalytic of version 12.2 sometimes produces incorrect results, e.g. FunctionAnalytic[Cos[Sqrt[z]], z] False FunctionAnalytic[Cos[Sqrt[z]], z, Complexes] FunctionAnalytic[Cos[Sqrt[z]],z,\[DoubleStruckCapitalC]] Up to Wiki…
user64494
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Split complex equation

Sometimes it is useful to split a complex equation into its real and imaginary parts. Consider the following ode ode = y'''[x] - k^3*y[x] == I*k*a*((2*x - c)*(y''[x] - k^2*y[x]) + 2*y[x]) bc1 = y'[0] == 0; bc2 = y''[1] + k^2*y[1]/(1 - c) == 0; in…
Nobody
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