ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Yoda on March 03, 2014, 08:15:15 pm
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One of the solutions to the equation z^12 = a; where a is a real number is, a^1/12cis(7pi/6). Find the number of solutions that have an imaginary part that is less than zero.
Anyone that can help me through this question thanks?
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If z^12=a ,you will have twelve solutions evenly spaced. 2pi / 12= pi/6 .each solution will be pi/6 away from the nearest solution. For imaginary part to be less than zero , your argument must be in 3rd or 4th quadrant and cannot be 0 or pi. I think there will be 5 solutions with imaginary part less than zero. It will be easier if your draw it out.
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There are actually an infinite number of solutions unless you limit the argument of the cis function, but assuming we are limited to
there are 5.
The equation 
has n solutions where 
: they are on the unit circle and evenly spaced at a distance of 
radians from each other.
To get the solutions with imaginary part less than zero look at the solutions on the bottom half of the unit circle: 
.
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There are actually an infinite number of solutions unless you limit the argument of the cis function, but assuming we are limited to there are 5.
The equation has n solutions where : they are on the unit circle and evenly spaced at a distance of radians from each other.
To get the solutions with imaginary part less than zero look at the solutions on the bottom half of the unit circle: .
Be careful here. Although there are infinitely many values for the unrestricted argument, these only correspond to a finite set of complex numbers as the cis function is periodic. For instance, you may get pi/12 and 23pi/12 as different arguments, but they correspond to the same complex number.
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Oh yeah, of course. For some reason I was just thinking about arguments the whole way through, as you can tell by my "solutions" ::)