How to convert complex number to trigonometric formula?
ā1 + iā 3
ANSWER : 2(cos(2Ļ/3) + i sin(2Ļ/3))
MATLAB: Making complex number to trigonometric formula
complex numbertrigonometric formula
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"The complex number is being squared and should result in a real answer with no complex part at all"
Huh? Certainly not. 
. You'll only get a real output if your input number is pure imaginary or pure real.
. You'll only get a real output if your input number is pure imaginary or pure real.So you can get complex numbers out of a square and as per Madhan's answer if you try to plot that, matlab will discard the imaginary part for the plot and warn you.
Now if the input to acos is pure real, then acos should return either a pure real or a pure complex number which when squared will be pure real, so it would seem that your input to acos is complex. You haven't supplied enough information for us to know why.
In any case, have you used the debugger to find out what is really happening with your code instead of what you think is happening?
First, you need to understand that sqrt(3)/2 is NOT 0.8660. There is a difference. Small errors like that will often be significant.
Second, you need to recognize that sin and cos are written to work in RADIANS. This is true of most programming languages. So, while 60 may mean 60 degrees to you, MATLAB will generally interpret it in the form of 60 radians.
So, IF 60 is intended to be in the form of radians, then we have:
sin(60)ans = -0.30481cos(60)ans = -0.95241However, since you seem to have substituted 1/2 for cos(60), I can only assume that you intended the number 60 to be in degrees.
Substituting the identity you used, we get:
soln = solve(2*cos(2*x) - 2*(cos(-2*x)*1/2 - sin(-2*x)*sqrt(3)/2) == 0,x)soln = -(log(- 3^(1/2)/2 - 1i/2)*1i)/2 -(log(3^(1/2)/2 + 1i/2)*1i)/2That may not look like 15 degrees to you, but...
rad2deg(double(soln))ans = -75 + 1.6214e-71i 15 + 1.6214e-71iFeel free to ignore the imaginary part, that is 71 powers of 10 smaller than the real part. It is zero.
So there is no mystery at all. Apparently the symbolic toolbox was intelligent enough that it saw your original equation with the number 60 inside a trig function, and realized that you must be working in degrees.
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