Question

Find all cube roots of the complex number in the image provided. Workout out the problem and leave the answers in polar form.

Answer 1

Given the complex number:

`z=r\cdot(\cos \theta+i\cdot\sin \theta)\text{.}`

The n n-th roots of the complex number z are given by:

`w_k=\sqrt[n]{r}\cdot\lbrack\cos ((\theta+k\cdot360\degree)/(n))+i\cdot\sin ((\theta+k\cdot360\degree)/(n))\rbrack\text{ where }k=0,1,2,\ldots,n-1.`

We must find all the cube roots of the following complex number:

`z=64\cdot(\cos (219\degree)+i\cdot\sin (219\degree)).`

For this number, we identify:

`\begin{gathered} r=64, \\ \theta=219\degree. \end{gathered}`

Using the formula above with these numbers, we get:

`\begin{gathered} w_0=\sqrt[3]{64}\cdot\lbrack\cos ((219\degree+0\cdot360\degree)/(3))+i\cdot\sin ((219\degree+0\cdot360\degree)/(3))\rbrack=4\cdot\lbrack\cos (73\degree)+i\cdot\sin (73\degree)\rbrack, \\ w_1=\sqrt[3]{64}\cdot\lbrack\cos ((219\degree+1\cdot360\degree)/(3))+i\cdot\sin ((219\degree+1\cdot360\degree)/(3))\rbrack=4\cdot\lbrack\cos (193\degree)+i\cdot\sin (193\degree)\rbrack, \\ w_2=\sqrt[3]{64}\cdot\lbrack\cos ((219\degree+2\cdot360\degree)/(3))+i\cdot\sin ((219\degree+2\cdot360\degree)/(3))\rbrack=4\cdot\lbrack\cos (313\degree)+i\cdot\sin (313\degree)\rbrack\text{.} \end{gathered}`

Answer

`\begin{gathered} w_0=4\cdot\lbrack\cos (73\degree)+i\cdot\sin (73\degree)\rbrack \\ w_1=4\cdot\lbrack\cos (193\degree)+i\cdot\sin (193\degree)\rbrack \\ w_2=4\cdot\lbrack\cos (313\degree)+i\cdot\sin (313\degree)\rbrack \end{gathered}`