Proposition 1.1.10: Two complex roots

Show that every equation z2 = c has exactly two solutions in C (unless c = 0). In other words, every complex (non-zero) number c has exactly two complex roots.
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We write, as usual, z = x + iy and expand the equation:

z2 = (x + iy)2 = x2 - y2 + 2ixy = a + ib

This results in a system of two equations:

(1)       x2 - y2 = a
(2)       2xy = b

Solving the second equation and substituting into the first gives

x2 - (b/2x)2 = a

or equivalently

x4 - ax2 - b2/4 = 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:

Interactive Complex Analysis, ver. 1.0.0
(c) 2006-2007, Bert G. Wachsmuth
Page last modified: May 29, 2007