find all the roots of z^5-i=0

Question

how do we solve this problem: Find all The roots of z^5-i=0 ?

 

Please HELP

Answer 1

zⁿ = rⁿ(cos nθ + i sin nθ), where n = power, r = modulus, θ = angle of radius.

 

Here n = 1/5, r = 1, θ = π/2

 

i^1/5 = [ 1^1/5 (cos θ + i sinθ)^1/5 ]

 

i^1/5 ∈ {(cis π/2)^1/5 , (cis 3π/2)^1/5 , (cis 5π/2)^1/5 , (cis 7π/2)^1/5 ,(cis 9π/2)^1/5 }

i^1/5 ∈ {(cis π/10),  (cis 3π/10),  (cis 5π/10),  (cis 7π/10),  (cis 9π/10) }

 

Note: all of these values are irrational.

Answer 2

When you say all the roots, I assume you mean the sq roots. So I'll give it a try.

 

(z^5-i)=0

 

First complete the square. This is done by multiplying by the opposite term. Example:

 

(x+3)=  (x+3)(x-3)= x^2+3x-3x-9= x^2-9

 

So do the same thing here.

 

(z^5-i)(z^5+i)=z^10+z^5i-z^5i-i^2=z^10-(-1)=z^10+1

 

z^10+1=0

 

Now take sqrt if that's the root they want solved, because there are an infinite number of roots.

 

Sqrt(z^10+1)=sqrt(z^10)+sqrt(1)= +/-(z^5+1)

 

So +(z^5+1) = (z^5+1) and -(z^5+1)=(-z^5-1)

 

Easier to just say +/-(z^5+1).  Most teachers will accept it inThais form.