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.Back
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. |
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:
