How do you find the complex roots of #t^4-1=0#?

1 Answer
Jan 7, 2017

Answer:

#t^4-1=0# has roots #+-1# and #+-i#

In a sense these are all complex roots, but #+-i# particularly so, in that they are not Real roots.

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We find:

#0 = t^4-1#

#color(white)(0) = (t^2)^2 - 1^2#

#color(white)(0) = (t^2 - 1)(t^2 + 1)#

#color(white)(0) = (t^2 - 1^2)(t^2 - i^2)#

#color(white)(0) = (t - 1)(t+1)(t- i)(t+i)#

Hence #t = +-1# or #t = +-i#

The roots #+-i# are non-Real Complex roots.

The roots #+-1# are Real roots. Since Real numbers are a subset of Complex numbers, #+-1# are also Complex roots.