[Math] Precalculus – Complex numbers/Roots of Unity/Exponential Form

complex numbersroots-of-unity

Find the roots of $6z^5 + 15z^4 + 20z^3 + 15z^2 + 6z + 1 = 0.$

Erm…it sorta looks like $(x+y)^6 = x^6 +6x^5y + 15x^4y^2 + 20x^3y^3 + 15y^4x^2 + 6y^5x + y^6$ Any idea how to proceed from here? Thanks for any help!

Best Answer

Add $z^6$ to both sides. Then the LHS is $(z + 1)^6$ . Now can you continue on your own ?

UPDATE:

After adding $z^6$ to both sides we have:

$$(z+1)^6 = z^6 \iff (z+1)^6(1 + 0i) = z^6$$

Now take 6th root from both sides and then we have:

$$(z+1)\sqrt[6]{1 + 0i} = z$$

Now actually the second factor represents the 6th root of unity. Convert it to polar form and then apply De Moivre's Formula to get sixth root. So we have:

$$\sqrt[6]{1 + 0i} = (\cos (2k\pi) + i \sin (2k\pi))^{\frac 16} = \cos \left(\frac{2k\pi}{6}\right) + i \sin \left(\frac{2k\pi}{6}\right)$$

Now plug any value for $k = 0,1,2,3,4,5$ and you'll get all six roots of unity. But we will have six roots, but a fifth degree polynomial have five. So one root doesn't provide a solution and that the root when $k=0$, because this will left us with $z + 1 = z$, which obviously isn't possible.

Now it's should be easy. Because the base angle in the polar form is $\frac{\pi}{3}$ we can get nice sine and cosine value.

Now I'll work through one example and leave the other four to you. For $k=1$ for the sixth root fo unity we have:

$$\sqrt[6]{1 + 0i} = \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) = \frac{1 + \sqrt{3}i}{2}$$

Now write $z$ as $z=a + bi$ and we need to solve:

$$(a + bi + 1)\frac{1 + \sqrt{3}i}{2} = a + bi$$

Multiply out and regroup:

$$a + bi + 1 + a\sqrt{3}i - b\sqrt{3} + \sqrt{3}i = 2a + 2bi$$ $$(a + 1 - b\sqrt{3}) + i(a\sqrt{3} + b + \sqrt{3}) = 2a + 2bi$$

Now we need to solve the following system of equations:

\begin{cases} a + 1 - b\sqrt{3} = 2a \\ a\sqrt{3} + b + \sqrt{3} = 2b \end{cases}

\begin{cases} a = 1 - b\sqrt{3} \\ (1 - b\sqrt{3})\sqrt{3} + b + \sqrt{3} = 2b \end{cases}

\begin{cases} a = 1 - b\sqrt{3} \\ \sqrt{3} - 3b + b + \sqrt{3} = 2b \end{cases}

\begin{cases} a = 1 - b\sqrt{3} \\ \sqrt{3} = 2b \end{cases}

\begin{cases} a = 1 - \frac{\sqrt{3}}{2}\sqrt{3} = 1 - \frac 32 = - \frac 12 \\ b = \frac{\sqrt{3}}{2} \end{cases}

So one root is $z = \frac{-1 + \sqrt{3}i}{2}$. Is ite clearer now? Can you do the same for other 4 cases?

Related Question