Find all solutions of the equation z.

Question

I have to solve $(z+1)^3=z^3$

First , from $(a+b)^3=a^3+3a^2b+3ab^2+b^3$

I have $z^3+3z^2+3z+1 = z^3$

Thus , $3z^2+3z+1 = 0$

From I have $\displaystyle\frac{-b\pm(b^2-4ac)^\frac{1}{2}}{2a} $

i.e. $\qquad\displaystyle\frac{(-3)\pm(9-12)^\frac{1}{2}}{6}=0$

i.e. $\qquad\displaystyle\frac{(-3)\pm(-3)^\frac{1}{2}}{6}$

Thus, I have $\displaystyle\frac{-3\pm\sqrt{3}i}{6}$

Therefore $\displaystyle z_1 = -\frac{1}{2}+\frac{\sqrt{3}i}{6}\,\,$ and $\displaystyle z_2 = -\frac{1}{2}-\frac{\sqrt{3}i}{6}$

Please check my solution , Thank you.

Answer 1

Plugging your $z_1$ in the equation, we get

$$\left(\frac{1+i}2\right)^3=\left(\frac{-1+i}2\right)^3,$$ which cannot be true.

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The solutions of

$$3z^2+3z+1=0$$ are

$$\frac{-3\pm\sqrt{9-12}}6=\frac{-3\pm i\sqrt3}{6}.$$

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For a more "trigonometric" solution,

$$\left(1+\frac1z\right)^3=1=e^{i2k\pi}$$

so that

$$z=\frac1{e^{i2k\pi/3}-1},$$

for $k=1,2$.

Answer 2

We have for $z\neq 0$

$$(z+1)^3=z^3 \iff \left(1+\frac1z\right)^3=1 \implies 1+\frac1z=1,e^{i\frac23 \pi},e^{-i\frac23 \pi}$$

therefore

$$\frac1z =-1+e^{i\frac23 \pi},-1+e^{-i\frac23 \pi} \implies z=\frac{\bar z}{z\bar z}=\frac1{2}(-1+e^{-i\frac23 \pi}),\frac1{2}(-1+e^{i\frac23 \pi})$$

that is

$$z_1= -\frac32-i\frac32 \sqrt 3, \quad z_2= -\frac32+i\frac32 \sqrt 3$$

Answer 3

Equivalently, you want to find all complex numbers $z$ for which $z$ and $z+1$ are cube roots of the same complex number $c$. In the Argand plane, the three cube roots of a complex number $c$ lie at the vertices of an equilateral triangle with centroid at the origin. (This is a fun exercise that is left to the reader!)

The numbers $z$ and $z+1$ are two vertices of an equilateral triangle if and only if the side length of the triangle is $1$ and one side is parallel to the horizontal axis (and bisected by the vertical axis). Therefore the real parts of $z$ and $z+1$ are $\pm{1\over2}$, so $z$ has real part $-{1\over2}$.

The distance between the centroid of an equilateral triangle and any side is one-third the triangle’s height, and the height of an equilateral triangle with side $1$ is $\sqrt3\over2$, so the complex part of $z$ (and of $z+1$) is $\pm\sqrt3\over2$, whence $z=-{1\over2}\pm i{\sqrt3\over6}$.