How do you find the complex roots of #t^4-1=0#?
1 Answer
Jan 7, 2017
In a sense these are all complex roots, but
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
The roots
The roots