## Proposition 1.1.10: Two complex roots |

Show that every equation |

We write, as usual, *z = x + iy* and expand the equation:

z^{2}= (x + iy)^{2}= x^{2}- y^{2}+ 2ixy = a + ib

This results in a system of two equations:

(1) x^{2}- y^{2}= a

(2) 2xy = b

Solving the second equation and substituting into the first gives

x^{2}- (b/2x)^{2}= a

or equivalently

x^{4}- ax^{2}- b^{2}/4 = 0

We can now apply the quadratic formula to find:

Since *x* (and *y*) must be real we have:

and consequently

To confirm that our calculations are correct you can verify that the two
square-roots of *1+i* are: