What is the fourth root of #-16# ?

2 Answers
Gaya3
Oct 30, 2017

#2sqrti#

Explanation:

#root4(-16)=root4(2*2*2*2*-1)#

#=2root4(-1)#

#=2*(-1)^(1/4)#

#=2*((-1)^(1/2))^(1/2)#

#=2*sqrt(sqrt(-1))#

#=2*sqrti#

George C.
Oct 30, 2017

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:

#-16# has four fourth roots, being the four roots of:

#x^4+16 = 0#

The principal fourth root is the one in Q1 with minimum positive #Arg# value.

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# to find the zeros:

#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 #-16# are:

#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 #sqrt(2)+sqrt(2)i#

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]}

Related questions
Impact of this question
2281 views around the world
You can reuse this answer
Creative Commons License