What is the fourth root of #-16# ?
2 Answers
Explanation:
Principal fourth root:
#root(4)(-16) = sqrt(2)+sqrt(2)i#
There are three other fourth roots of the form:
#+-sqrt(2)+-sqrt(2)i#
Explanation:
#x^4+16 = 0#
The principal fourth root is the one in Q1 with minimum positive
Use:
#a^2-b^2 = (a-b)(a+b)#
#a^4+b^4 = (a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)#
Let's factorise
#x^4+16#
#= (x^2-2sqrt(2)x+4)(x^2+2sqrt(2)x+4)#
#= ((x-sqrt(2))^2-(sqrt(2)i)^2)((x+sqrt(2))^2-(sqrt(2)i)^2)#
#= ((x-sqrt(2))-sqrt(2)i)((x-sqrt(2))+sqrt(2)i)((x+sqrt(2))-sqrt(2)i)((x+sqrt(2))+sqrt(2)i)#
#= (x-sqrt(2)-sqrt(2)i)(x-sqrt(2)+sqrt(2)i)(x+sqrt(2)-sqrt(2)i)(x+sqrt(2)+sqrt(2)i)#
So the four fourth roots of
#sqrt(2)+sqrt(2)i#
#sqrt(2)-sqrt(2)i#
#-sqrt(2)+sqrt(2)i#
#-sqrt(2)-sqrt(2)i#
The principal one, which is in Q1, is
Here are the roots plotted in the complex plane:
graph{((x-sqrt(2))^2+(y-sqrt(2))^2-0.002)((x-sqrt(2))^2+(y+sqrt(2))^2-0.002)((x+sqrt(2))^2+(y-sqrt(2))^2-0.002)((x+sqrt(2))^2+(y+sqrt(2))^2-0.002) = 0 [-5, 5, -2.5, 2.5]}