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2nd degree equation (complex numbers)

  1. Aug 22, 2007 #1
    I just came across the eq.

    z^2 - 2z + 1 - 2i

    where z is a complex number. How do I solve this sort of eq.?

    I tried to solve it as a normal 2nd degree eq., setting a=2, b=-2 and c=(1-2i), with z as the variable. This finally gave me the solutions

    z(1) = -1 + sqrt(2i)

    and

    z(2) = -1 - sqrt(2i)

    Can this be the correct solution? I had hoped for an answer involving i, not sqrt(i)...
     
  2. jcsd
  3. Aug 22, 2007 #2
    This is correct -- though you might want to find the solution to [tex]a+b i = \sqrt{i}[/tex] where a and b are real numbers. Remember that the point of the complex numbers to to be closed under things like taking square roots.
     
  4. Aug 22, 2007 #3
    A relevant equation might be
    [tex]\left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right)^2 = i[/tex]

    which can be found graphically by viewing the unit circle [tex]|z|=1[/tex] in the complex plane, and considering that if [tex]z = |z|_{\theta}[/tex], then [tex]z^n = |z|^n _{n \theta}[/tex], where [tex]\theta[/tex] is the angle between the lines corresponding to [tex]z[/tex] and [tex]1+0i[/tex].
     
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