Question

find all fifth roots of 4-4i​

Answer 1

Step-by-step explanation:

The complex 5th roots of −4+4i are given by:

(−4+4i)1/5={2–√cis(27+72k)∘}

for k=0,1,2,3,4.

That is:

k=0: z=z1 = 2–√cis27∘

k=1: z=z2 = 2–√cis99∘

k=2: z=z3 = 2–√cis171∘

k=3: z=z4 = 2–√cis243∘

k=4: z=z5 = 2–√cis315∘

Proof

Complex 5th Roots of -4 + 4i.png

Let z5=−4+4i.

We have that:

z5=42–√cis(3π4+2kπ)=42–√cis(135∘+k×360∘)

Let z=rcisθ.

Then:

z5 = r5cis5θ De Moivre's Theorem

= 42–√cis(3π4+2kπ)

⇝ r5 = 42–√

5θ = 3π4+2kπ

⇝ r = (42–√)1/5

= 2–√

θ = 3π20+2kπ5 for k=0,1,2,3,4

= 27∘+72k∘ for k=0,1,2,3,4

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