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Showing content from https://en.cppreference.com/w/cpp/language/../algorithm/../ranges/../numeric/complex/pow.html below:

std::pow(std::complex) - cppreference.com

(1) (until C++11) (2) (3) (4) (A) (until C++23) (since C++23) (B) template< class T, class NonComplex > std::complex</* common-type */> pow( const std::complex<T>& x, const NonComplex& y ); (until C++23) (since C++23) (C) template< class T, class NonComplex > std::complex</* common-type */> pow( const NonComplex& x, const std::complex<T>& y ); (until C++23) (since C++23)

1-4) Computes complex x raised to a complex power y with a branch cut along the negative real axis for the first argument. Non-complex arguments are treated as complex numbers with positive zero imaginary component.

A-C)

Additional overloads are provided.

NonComplex

is not a specialization of

std::complex

(since C++11) [edit] Parameters [edit] Return value

1-4) If no errors occur, the complex power xy, is returned.

Errors and special cases are handled as if the operation is implemented by

std::exp(y * std::log(x))

The result of

std::pow(0, 0)

is implementation-defined.

A-C) Same as (2-4).

[edit] Notes

Overload (1) was provided in C++98 to match the extra overloads (2) of std::pow. Those overloads were removed by the resolution of LWG issue 550, and overload (1) was removed by the resolution of LWG issue 844.

The additional overloads are not required to be provided exactly as (A-C). They only need to be sufficient to ensure that for their first argument base and second argument exponent:

If base and/or exponent has type std::complex<T>:

If base and/or exponent has type std::complex<long double> or long double, then std::pow(base, exponent) has the same effect as std::pow(std::complex<long double>(base), std::complex<long double>(exponent)).
Otherwise, if base and/or exponent has type std::complex<double>, double, or an integer type, then std::pow(base, exponent) has the same effect as std::pow(std::complex<double>(base), std::complex<double>(exponent)).
Otherwise, if base and/or exponent has type std::complex<float> or float, then std::pow(base, exponent) has the same effect as std::pow(std::complex<float>(base), std::complex<float>(exponent)).

(until C++23)

If one argument has type std::complex<T1> and the other argument has type T2 or std::complex<T2>, then std::pow(base, exponent) has the same effect as std::pow(std::complex<std::common_type_t<T1, T2>>(base), std::complex<std::common_type_t<T1, T2>>(exponent)).

If std::common_type_t<T1, T2> is not well-formed, then the program is ill-formed.

(since C++23) [edit] Example

#include <complex>
#include <iostream>
 
int main()
{
    std::cout << std::fixed;
 
    std::complex<double> z(1.0, 2.0);
    std::cout << "(1,2)^2 = " << std::pow(z, 2) << '\n';
 
    std::complex<double> z2(-1.0, 0.0); // square root of -1
    std::cout << "-1^0.5 = " << std::pow(z2, 0.5) << '\n';
 
    std::complex<double> z3(-1.0, -0.0); // other side of the cut
    std::cout << "(-1,-0)^0.5 = " << std::pow(z3, 0.5) << '\n';
 
    std::complex<double> i(0.0, 1.0); // i^i = exp(-pi / 2)
    std::cout << "i^i = " << std::pow(i, i) << '\n';
}

Output:

(1,2)^2 = (-3.000000,4.000000)
-1^0.5 = (0.000000,1.000000)
(-1,-0)^0.5 = (0.000000,-1.000000)
i^i = (0.207880,0.000000)

[edit] See also

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