Inverse cosine of complex numbers

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Sriram 2014-10-9

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编辑： Andrew Reibold 2014-10-10

I'm trying to get the inverse cosine of a complex number. Using the definition given in the help of acos(z) produces a different result than directly using acos. See below for the code snippet. Can someone help me out?

format LONGENG

 z = 0.625820036622262+1.576125391478034E-055i;
 res1 = acos(z)
d = z + i*sqrt(1-(z*z))
res2 = -i*log(d)

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回答（1 个）

Andrew Reibold 2014-10-9

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编辑：Andrew Reibold 2014-10-9

在 MATLAB Online 中打开

I know why!

You were close, but you didn't quite replicate the definition in the documentation correctly.

This is how you would write it.

d = z + i*sqrt(1-z*z)
res2 = -i*log(d)

Using this corrected definition, I get approximately 0.894613863 + ~0i for both answers.

First Method Solution: 0.894613863926113 -  2.02075780910344e-55i
Second Method Solution: 0.894613863926114 +  1.11022302462516e-16i

Bother of the i values are near negligible in magnitude relative to the real part.

Hope this helped!

-Andrew

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Sriram 2014-10-9

Can we know how acos is implemented within MATLAB for them to get -2.02075780910344e-55i? Clearly it must some numerical implementation -

Andrew Reibold 2014-10-10

编辑：Andrew Reibold 2014-10-10

For some reason I thought your square root function was only taking the square root of z^2. Unless you edited the question, I just made an honest mistake saying you typed it wrong I guess.

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