## Example 1.1.5 (b): Properties of i |

Use the definition of |

We already know that *z=i* solves *z ^{2}+1=0*.
But if

z^{2}=(-i)*(-i) = (-1)^{2}i^{2}= -1

Thus, *z=-i* is a second solution. These are the only two solutions,
because

z^{2}+1 = (z+i)(z-i)

To find all solutions to *x ^{4}+1=0* we first note that

z^{4}+1 = (z^{2}+i)(z^{2}-i)

Thus, we need to find the roots of *i* and *-i* next.
In other words, we need to find a complex number *z = x+iy* such
that

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

Lokking at the real and imaginary parts separately, this results in two equations:

(1) x^{2}-y^{2}= 0

(2) 2xy = 1

From there, and using a similar system to solve *z ^{2} = -i*
we get the following four solutions to

z_{1}= 1/(1+i)

z_{2}= 1/(-1+i)

z_{3}= 1/(-1-i)

z_{4}= 1/(1-i)

Later it will turn out that any *n*-th degree polynomial has
*n* roots, and they sometimes have some nice geometric properties.