# Solving cube complex equation with z and i

I'm trying to solve the below equation

$$(z-1)^3 - (1-i)(z+1)^3 = 0$$

I tested on two approaches:

1) Doing some operations, the above equation becomes:

$$-6z^2 - 2 + (z+1)^3i = 0$$

So, I consider the first part as real part and the second one as imagine part and I tried to solve the equations. Particularly,

$$-6z^2 - 2 = 0$$ and $$(z+1)^3 = 0$$

The roots are $$- \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3} , -1$$

2) Doing also some operations, but the results are different,

$$(\frac{z-1}{z+1})^3 = (1-i) \rightarrow ... \rightarrow z = \frac{1 + \sqrt[3]{1-i}}{1 - \sqrt[3]{1-i}}$$

So, my questions are. Which of the above two approaches is correct? Is there any different way to solve this problem ? I am very confused because there is $$z$$ in the equation with $$i$$ and I don't know how can I deal with it.

The first approach is not correct. You can have $$i(z+1)^3-(6z^2+2)=0$$ without having $$(z+1)^3=0$$ and $$6z^2=0$$.

The other approach is fine. Now, use the fact that $$1-i=\sqrt2e^{-\pi i/4}$$ to deduce that the cube roots of $$1-i$$ are$$\sqrt[6]2e^{-\pi i/12},\ \sqrt[6]2e^{7\pi i/12}\text{ and }\sqrt[6]2e^{5\pi i/4}.$$

• how do I know that the $1-i = \sqrt{2}e^{-\pi i/4}$ ? Thank you for your answer – Dimitris Dimitriadis Sep 25 at 12:28
• Because$$1-i=\sqrt2\left(\frac1{\sqrt2}-\frac i{\sqrt2}\right)=\sqrt2\left(\cos\left(-\frac\pi4\right)+\sin\left(-\frac\pi4\right)i\right).$$ – José Carlos Santos Sep 25 at 12:37

Your first approach is wrong. $$z$$ is not going to be real, so there's no reason to think that the real and imaginary parts would be those. Your second approach is correct, but there are three complex cube roots of $$1-i$$.

Because $$z$$ need not be real, you can't assume that $$-6z^2-2$$ is the real part of your equation. Note that if you try to solve $$6z^2=-2$$ you are taking the square root of a negative number and should get an imaginary value for $$z$$.

In the second (correct) method you need to make sure that you take just one consistent value of the cube root (there are three possibilities representing the three roots of he original cubic). You can compute this explicitly using $$1-i=re^{i\theta}$$

Explicit roots of $$z$$

$$\left(\frac{z-1}{z+1} \right)^3=1-i=2^{1/2} e^{i\pi(2n-1/4)} \Rightarrow \left(\frac{z-1}{z+1} \right)=2^{1/6} e^{i\pi(2n-1/4)/3}=t_n, n=0,1,2.$$ Then $$z_n=\frac{1+t_n}{1-t_n}, t_n,~ n=0,1,2$$