C# Math.Pow(-2,1.1) > n.def

Question

I just found out that when entering a negative x and a decimal y, Math.Pow() returns the not-defined value as result, which is wrong I guess. Calculating this in other programs, even like the windows Calculator works with a correct result. Also this case is not mentioned in the documentation.

Target Framework is 4.

Can anyone explain this?

Answer 1

The result is going to be complex number, so you have to use Complex class from System.Numerics namespace.

Complex n = new Complex(-2, 0);
Complex result = Complex.Pow(n, 1.1);

In case if result is real number (integer power), then you can use Math.Pow.

As @JeppeStigNielsen mentioned, the conversion from int/double to Complex is implicit, so the code can be shortened to:

Complex result = Complex.Pow(-2, 1.1);

Answer 2

Also this case is not mentioned in the documentation

You sure? From it's documentation;

Parameters

x < 0 but not NegativeInfinity; y is not an integer, NegativeInfinity, or PositiveInfinity.

Return value

NaN

I'm not sure which OS you tried but it doesn't work in calc.exe (Win7 - 64bit) says Invalid Input.

As Ulugbek mentioned, taking 1.1 power of a negative value creates a complex number. Because (-2)^1.1 = (-2)^11/10 = (-2)^1/1011 and getting 10 times rooth of -2 returns a complex number.

Since Math.Pow takes and returns double values, this doesn't fit with requirements. You can use Complex class from System.Numerics.

http://www.wolframalpha.com/input/?i=-2^1.1

Further reading: How is Math.Pow() implemented in .NET Framework?