Example 1.1.5 (b): Properties of i

Use the definition of i to solve z2+1=0 and z4+1=0
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We already know that z=i solves z2+1=0. But if z=-i, then

z2 =(-i)*(-i) = (-1)2 i2 = -1

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

z2+1 = (z+i)(z-i)

To find all solutions to x4+1=0 we first note that

z4+1 = (z2+i)(z2-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

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

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

(1)       x2-y2 = 0
(2)       2xy = 1

From there, and using a similar system to solve z2 = -i we get the following four solutions to z4=-1:

z1 = 1/(1+i)
z2 = 1/(-1+i)
z3 = 1/(-1-i)
z4 = 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.

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