Problem with the sqrt function

3 次查看（过去 30 天）

Chirag Anand 2017-6-8

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/343868-problem-with-the-sqrt-function

评论： Stephen23 2017-6-8

when I use the sqrt function to find the sqrt of a determinant of a complex matrix, sometimes matlab changes the sign of the real and imaginary part of the square root. for example if the it is -x+iy the sqrt function gives the answer as x-iy How do I get around this?

2 个评论
显示无隐藏无

Adam 2017-6-8

编辑：Adam 2017-6-8

在 MATLAB Online 中打开

Just like in the real case, if a is a square root then so is -a. In the real case the positive square root is always returned, for complex numbers the documentation states that sqrt(z) returns:

sqrt(r)*(cos(phi/2) + 1i*sin(phi/2))

where r = abs(z) is the radius and phi = angle(z) is the phase angle on the closed interval -pi <= phi <= pi.