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Math Help - Finding the roots of a complex number

  1. #1
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    Finding the roots of a complex number

    The question:
    Find the roots (in Cartesian Form) of z^2 - 3z + (3 - i) = 0

    I'm not sure where to start. I tried applying the quadratic formula, but it doesn't seem to help. Any help would be great.
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  2. #2
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    By the quadratic formula you should get

    [LaTeX ERROR: Convert failed]

    From here you need to use that the square root is multivalued.
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  3. #3
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    It is simplest to take roots by using DeMoivre's formula: The nth roots of complex number z= re^{i\theta}= r(cos(\theta)+ i sin(\theta)), in polar form, are given by \sqrt[n]{z}= r^{1/n}e^{i\frac{\theta+ 2k\pi}{n}}= r^{1/n}(cos(\frac{\theta+ 2k\pi}{n})+ i sin(\frac{\theta+ 2k\pi}{n})) where k= 0 to n-1 give the n roots. (r must be positive of course, and r^{1/n} is the positive real n th root.)

    In particular, if [itex]a+ bi= r e^{i\theta}= r(cos(\theta)+ i sin(\theta))[/itex] then the square root of a+ bi is given by \sqrt{r}e^{i\frac{\theta}{2}}= \sqrt{r}(cos(\frac{\theta}{2})+ i sin(\frac{\theta}{2})) and \sqrt{r}e^{i\frac{\theta}{2}}= \sqrt{r}(cos(\frac{\theta}{2}+ \pi)+ i sin(\frac{\theta}{2}+ \pi))
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  4. #4
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    Quote Originally Posted by Glitch View Post
    The question:
    Find the roots (in Cartesian Form) of z^2 - 3z + (3 - i) = 0

    I'm not sure where to start. I tried applying the quadratic formula, but it doesn't seem to help. Any help would be great.
    An alternative approach to the excellent approaches already suggested is to substitute z = a + ib, expand and equate the real and imaginary parts to zero. This gives two simultaneous equations to be solved for a and b, leading to the two solutions z = 2 + i and z = 1 - i.
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  5. #5
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    Quote Originally Posted by Vlasev View Post
    By the quadratic formula you should get

    [LaTeX ERROR: Convert failed]

    From here you need to use that the square root is multivalued.
    OK, I got this far. I'm not sure where I'm supposed to go from here. Is it a case of working out the square root using De Moivre's theorem, by converting to exponential form first?
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  6. #6
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    Yea. You can do that. You will get two solutions (1+2i) and (-1-2i). Plugging both in the formula above, you will get a total of 2 different solutions.
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