Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Square Root of Complex Numbers

  1. Dec 25, 2006 #1
    Hi! I've got a question.
    There is a nice formula for finding square roots of arbitrary complex numbers z=a+bi:

    [itex]\frac{1}{\sqrt{2}}(\epsilon\sqrt{|z|+a}+i\sqrt{|z|-a})[/itex] where
    epsilon:=sing(b) if b≠0 or epsilon:=1 if b=0.

    I've just looked it up and it's nice to use it to find complex roots of quadratic equations with complex coefficients.

    Where does it come from? I mean, squaring shows that it's true but how can one derive it from other facts? Is there a similar formula for n-th roots (not in polar form but analogous to that above)? Any info, links?
    Last edited: Dec 25, 2006
  2. jcsd
  3. Dec 26, 2006 #2

    Let y=c+id, where y^2=z.

    Then; c^2-d^2+2i c*d=a+ib.

    As a result; a=c^2-d^2 (eq1) and b=2c*d (eq2).

    Therefore, from eq1 and eq2; c^2 = a + (b/c)^2/4 (eq3);

    Solving eq3 in c^2, we get c^2={a+sqr(a^2+b^2)}^.5 /2

    Therefore, c=(+/-) sqr({a+mod(z)}/2) (eq4 ),
    {the other solution is rejected as c must be real}.

    Then; z^1/2 = (+/-) y = (+/-) [c + id] = (+/-) [c+id] = (+/-) [sqr({a+mod(z)})+i sign(b)* sqr({[mod(z)]^2 –a^2} / {a+mod(z)})]/sqr(2)
    = (+/-) [ sqr( a+mod(z) ) + i sign(b) * sqr({mod(z)–a} ] / sqr(2)
    = (+/-) [sign(b) * sqr( a+mod(z) ) + i sqr({mod(z)–a} ] / [sqr(2)* sign(b)]
    ====> (+/-) [sign(b) * sqr( a+mod(z) ) + i sqr({mod(z)–a} ] / [sqr(2)]

    Solving these equaitons when b=0 will need some modifications. By the way, when b is zero, z is real.

    Amr Morsi.
  4. Dec 26, 2006 #3
    It can probably be derived with few identities and Demoivre's theorem.
  5. Dec 26, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper

    You could always convert the number to e^(x + iy) form, divide x and y by 2, convert back to normal form. I think that should work.
  6. Dec 27, 2006 #5


    User Avatar
    Science Advisor
    Dearly Missed


    When you multiply by a complex number, you are rotating the other number
    by the angle the complex number makes with the x-axis.

    For example, multiplication by "i" makes a rotation of 90 degrees. Therefore
    a multiplication by the square of "i" makes a rotation of 180 degrees or -1.
    Hence "i" is the square root of -1.

    If you want the "n-th root" of a number; express it in polar form, and
    divide the angle by "n". You will get the polar expression for the n-th
    root. Use trigonometric identities to transform back into the non-polar
    form in terms of the components of the original complex number.
    Normalize appropriately if the modulus of the number is not unity.

    That's how the formula you cite above is derived.

    Dr. Gregory Greenman
  7. Dec 27, 2006 #6
    the question is..take the identity:

    [tex] (\sqrt (-2)+1)(\sqrt (-2)-1)=-2 [/tex] and expand it by a continous fraction..what would we get..
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook