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Problem with the sqrt function

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Chirag Anand
Chirag Anand 2017 年 6 月 8 日
コメント済み: Stephen23 2017 年 6 月 8 日
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?
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Adam
Adam 2017 年 6 月 8 日
編集済み: Adam 2017 年 6 月 8 日
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
Stephen23 2017 年 6 月 8 日
"Problem with the sqrt function"
What is the problem? sqrt is giving a correct output.

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