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Gib Z
Sep30-07, 02:54 AM
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P: 3,352
Unfortunately no, there is 1 method but it is only for square roots (it can be extended to cube and 4th roots only if you know how to solve general polynomials of degree 3 and 4, which is general is a very laborious and difficult process).

We have the knowledge that the complex field is closed ie Any algebraic operation applied on a complex number will yield another complex number.

With this knowledge we can conclude that for any complex number, x+iy , its square root will be some other complex number, let it be of the form a+bi. We begin the method thus:

[tex] \sqrt{ x+ iy} = a+bi[/tex]
[tex]x+iy = (a+bi)^2 = (a^2-b^2) + 2ab i[/tex]

By comparing co efficients of real and imaginary parts, we have a pair of simultaneous equations:

[tex]a^2-b^2 = x[/tex]

In the second equation, divide through by 2a to find an expression for b.

Substituting that into the first simultaneous equation gives:
[tex]a^2 - \frac{y^2}{4a^2} =x[/tex]

Now multiply through by [itex]4a^2[/itex] and equate the expression to 0.

Then we have a quadratic equation in a^2. Using the quadratic formula or otherwise, solve for a^2. When we take the square root of that, we have to keep both roots, the positive and the negative, because we are not sure which one is the correct one, or in fact both may me (remember there are always 2 square roots).

So you have 2 values of a. Each one yields a value of b when put back into either of the simultaneous equations, so there are 2 values of b as well.

So now you have 4 complex numbers, 2 of them being roots and 2 of them being introduced errors. You only need to check a single one, since the 4 roots come in 2 pairs, that are positive and negatives of each other. It should appear as below:

[tex] [a+bi, -a-bi] [-a+bi, a-bi][/tex].

To test one just take the number and square it. If it is correct, so is its negative partner. If its not correct, the other group is correct.

As you can see, you would probably want to do Polar form instead =D