Problem with the sqrt function
4 次查看(过去 30 天)
显示 更早的评论
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
Adam
2017-6-8
编辑:Adam
2017-6-8
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.
Stephen23
2017-6-8
"Problem with the sqrt function"
What is the problem? sqrt is giving a correct output.
回答(0 个)
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!