Find all solutions to the equation $(z+1)^6=z^6$. Express the solutions in simplified rectangular form. Hint: Clearly $z=0$ is not a solution.

Question

$$(z+1)^6=z^6 $$ $$\frac{(z+1)^6}{z^6 }=1$$ $$\left(\frac{(z+1)}{z }\right)^6=1$$ $$\left(\frac{(z+1)}{z }\right)^6=\cos(2*\pi*k)+i\sin(2*\pi*k)$$ $$\frac{(z+1)}{z }=\left(\cos(2*\pi*k)+i\sin(2*\pi*k)\right)^{1\over6}$$ $$1+\frac{1}{z }=\cos({2*\pi*k\over6})+i\sin({2*\pi*k\over6})$$ $$\frac{1}{z }=(\cos({\pi*k\over3})-1)+i\sin({\pi*k\over3})$$ $$\frac{1}{(\cos({\pi*k\over3})-1)+i\sin({\pi*k\over3}) }=z$$ I dont know how else to solve this. A lot of help is appreciated. Also i know that k should go from 1-5.

Answer 1

Hint: Solve $z+1 = \zeta_6^k z$ for $k = 0, \ldots, 5$, where $\zeta_6 = e^{\frac{i\pi}{3}}$.

Answer 2

Given the equation $z_{1}^{n}=z_{2}$, where $z_{1}=re^{i\theta}$ and $z_{2}=\rho e^{i\phi}$, the solution is given by $$z_{1}=\rho^{1/n}e^{i\frac{\phi+2\pi k}{n}}\,k\in\mathbb{Z}.$$

Let $w=\frac{z+1}{z}$. Then, $w^{6}=1$. Here, $1=1e^{i\cdot0}$, giving us $$w=e^{i\frac{0+2\pi k}{6}}=e^{i\frac{\pi k}{3}},\,k\in\mathbb{Z}$$ so $$\frac{z+1}{z}=e^{i\frac{\pi k}{3}},\,k\in\mathbb{Z}.$$

Solving for $z$, \begin{align*} z+1=ze^{i\frac{\pi k}{3}}\implies&z-ze^{i\frac{\pi k}{3}}=-1\implies z\left(1-e^{i\frac{\pi}{k}}\right)=-1\\ \implies&z=\frac{-1}{1-e^{i\frac{\pi k}{3}}},\,k\in\mathbb{Z}. \end{align*}

Answer 3

Since $(z+1)^6=z^6$, $\lvert z+1\rvert^6=\lvert z\rvert^6$, and therefore $\lvert z+1\rvert=\lvert z\rvert$. So, $z$ is equidistant from $-1$ and $0$; and therefore it can be written as $-\frac12+xi$, for some real number $x$. So, you solve the equation$$\left(\frac12+xi\right)^6=\left(-\frac12+xi\right)^6,$$which is equivalent to$$-6 i x^5+5 i x^3-\dfrac{3ix}8=0.\tag1$$Of course, $0$ is a root. Diving $(1)$ by $-6xi$, you get the equation $x^4-\dfrac{5x^2}6+\dfrac1{16}=0$, which is biquadratic and therefore easy to solve. Its roots are $\pm\dfrac{\sqrt3}6$ and $\pm\dfrac{\sqrt{3}}2$.