4th roots of complex number

[snozzla]

Would really appreciate help, I cannot simplify this complex number so i find modulus and arg of z so I can use De Moivre's formula.

Find the complex 4th roots of z:

z^1/4 = (-32/1+(squareroot of 3)*i)^1/4

 

[soroban]

Hello, snozzla!

\(\displaystyle \text{Find the complex 4th roots of: }\;z \:=\: \frac{-32}{1+i\sqrt{3}}\)

We need to rationalize that complex number.

\(\displaystyle \text{Multiply by }\frac{1-i\sqrt{3}}{1-i\sqrt{3}}: \;\;\frac{-32}{1+i\sqrt{3}}\cdot \frac{1-i\sqrt{3}}{1-i\sqrt{3}} \;=\;\frac{-32(1-i\sqrt{3})}{1 - 3i^2} \;=\;\frac{-32(1-i\sqrt{3})}{4}\)

. . \(\displaystyle \text{Hence: }\;z \;=\;-8 + 8\sqrt{3}\,i\)


\(\displaystyle \text{On the Argand diagram, we have:}\)

                      |
              *       |
              |\      |
              | \     |
              |  \    |
            _ |   \   |
          8/3 |    \  |
              |     \ |
              |      \|
      - - - - + - - - + - - -
                 -8   |

\(\displaystyle \text{And we find that: }\:r = 16,\;\theta \,=\,\tfrac{2\pi}{3}\)

\(\displaystyle \text{Can you continue?}\)

 

[snozzla]

Thank you very much! I will try to continue now

Found that

z0 = squareroot of 3 + 1/2 i
z1 = -1 + ((squareroot of 3)/2) * i

hopefully they are right