complex-numbers

问题 I am writing a program in Processing that transforms complex numbers. However, I want to have a method of taking an input string and calculating the transformation using a complex variable. For example: 1/(z+1) (z^2)/(z/2) where z is a complex number. Now, I've looked at JEP and some examples, but I cannot work out if it would allow you to actually enter z as a variable (and in any case it is not free). Is there an expression parser for Java (that works in processing, which uses an old

问题 I understand that domain or color wheel plotting is typical for complex functions. Incredibly, I can't find a million + returns on a web search to easily allow me to reproduce some piece of art as this one in Wikipedia: There is this online resource that reproduces plots with zeros in black - not bad at all... However, I'd like to ask for some simple annotated code in Octave to produce color plots of functions of complex numbers. Here is an example: I see here code to plot a complex function.

问题 I understand that domain or color wheel plotting is typical for complex functions. Incredibly, I can't find a million + returns on a web search to easily allow me to reproduce some piece of art as this one in Wikipedia: There is this online resource that reproduces plots with zeros in black - not bad at all... However, I'd like to ask for some simple annotated code in Octave to produce color plots of functions of complex numbers. Here is an example: I see here code to plot a complex function.

问题 take a look at the code blew: #include <complex> #include <iostream> int main() { std::cout << std::pow( std::complex<double>(0,0), std::complex<double>(0,0) ) << "\n"; std::cout << std::pow( std::complex<double>(0,0), double(0) ) << "\n"; return 0; } g++(4.8.1) gives an output of (nan,nan) (-nan,-nan) while clang++(3.3) gives an out put of (-nan,-nan) (-nan,-nan) But I am expecting (1.0, 0.0). Can anyone give an explanation? 回答1: According to std::pow documentation Return value base raised

问题 Using the complex class and library, how do I assign a complex number to a variable? I understand that I can set the value when I first instantiate the complex number. I also understand that I can assign one instantiated complex number to another. How can I directly assign a complex number to a variable? Reference: http://www.cplusplus.com/reference/complex/complex/operators/ Example: #include <iostream> #include <complex> int main() { complex<double> a(1.2,3.4), b; cout << a; //-> (1.2,3.4)

问题 I'm currently learning about discret Fourier transform and I'm playing with numpy to understand it better. I tried to plot a "sin x sin x sin" signal and obtained a clean FFT with 4 non-zero points. I naively told myself : "well, if I plot a "sin + sin + sin + sin" signal with these amplitudes and frequencies, I should obtain the same "sin x sin x sin" signal, right? Well... not exactly (First is "x" signal, second is "+" signal) Both share the same amplitudes/frequencies, but are not the

问题 The following MATLAB code snippet, which creates two arrays of complex numbers, x = complex(1:2,0:1); y = complex(zeros(1,2),0); whos x y prints Name Size Bytes Class Attributes x 1x2 32 double complex y 1x2 32 double complex as expected. However, after these two additional statements, y(1) = x(1); whos x y the following gets printed: Name Size Bytes Class Attributes x 1x2 32 double complex y 1x2 16 double How can one prevent the complex attribute from being dropped? For the record, Octave

问题 The following code with VS2010 prints 0 , contrary to my expectations: #include <complex> #include <iostream> using namespace std; int main(void) { complex<int> z(20, 200); cout << abs<int>(z) << endl; return 0; } It works fine when the type is double . 回答1: According to the C++ ISO spec, §26.2/2: The effect of instantiating the template complex for any type other than float , double or long double is unspecified. In other words, the compiler can do whatever it wants to when you instantiate

问题 i ran into something interesting about the python augmented assignment += it seems to be automatic data type conversion is not always done for a += b if a is a 'simpler' data type, while a = a + b seems to work always cases where the conversion is done a = 1 b = 1j a = 1 b = 0.5 case where the conversion is not done from numpy import array a = array([0, 0 ,0]) b = array([0, 0, 1j]) after a += b , a remains as integer matrix, instead of complex matrix i used to think a += b is the same as a =

问题 For a project in one of my classes we have to output numbers up to five decimal places.It is possible that the output will be a complex number and I am unable to figure out how to output a complex number with five decimal places. For floats I know it is just: print "%0.5f"%variable_name Is there something similar for complex numbers? 回答1: For questions like this, the Python documentation should be your first stop. Specifically, have a look at the section on string formatting. It lists all the