# For complex $z$, find all solutions to: $(z - 6 + i)^3 = -27$

For complex $z$, find all solutions to: $(z - 6 + i)^3 = -27$

I reasoned that for this to be true, $z - 6 + i$ must be $= -3$

$\therefore z - 3 + i = 0$

$z = 3 - i$

However, there are two more solutions provided for this problem and I am not sure how to get to them. What is the standard method for solving something like this to ensure that all answers are produced?

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The equation $u^3=-27$ has three solutions $u$ in $\mathbb C$. Can you describe them? Hint: polar representation of complex numbers and de Moivre's formula.

Once this is done, you will see that the equation $(z-6+\mathrm i)^3=-27$ has three solutions as well, and you will be able to write them down (one of them is $3-\mathrm i$, of course).

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$(z - 6 + i)^3 = -27$ is of the form $u^3=(-3)^3$, which can be factored over $\mathbb{C}$ as:

$$u^3-(-3)^3=(u-(-3))(u^2+u(-3)+(-3)^2)=(u+3)(u^2-3u+9)$$

One root is $u=-3$.

The other two roots, one can get after solving quadratic equation $u^2-3u+9=0$.

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First Try to find all the cube roots of $-27$.(Or whatever it is that is relevant to your problem)

To do this you write down Complex numbers in their Polar form. Or in the form
$$z=r(\cos\theta+\text{i}\sin\theta)$$ Where $\theta$ is the argument of $z$ and $r$ its modulus or norm (or its length (on the Argand Diagram) in simple terms). Finding out the cube roots of $-27$ is not a very difficult thing to do. But for this you require the knowledge of de Moivre's Theorem. It deals with computing powers of complex numbers. It says (for now, just trust the equation and move along:) I will not give you a proof here) $$r(\cos\theta+\text{i}\sin\theta)^n=r^n(\cos n\theta+\text{i}\sin n\theta)\tag{1}$$ How do you use this to find the cube roots? Well, Just assume that for some $u\in \mathbb C$ with argument $\alpha$ and modulus $q$, $$[q(\cos\alpha+\text{i}\sin\alpha)]^3=-27$$ Or, from the result of de Moivre's Theorem, $$q^3(\cos 3\alpha+\text{i}\sin 3\alpha)=-27(\cos 0+\text i \sin0)\tag{2}$$ Equation $(2)$ suggests that
$q^3=-27 ~~~~~\implies~ q=\sqrt[3]{-27}=-3$, and
$3\alpha=0~~~~~~~~~\implies3\alpha=0, 2\pi ~\text{or}~ 4\pi\implies~\alpha =0 ,~ \frac23\pi~~\text{or} ~~\frac43\pi~$

You now have three sets of possible values for $\alpha$ which is, if you remember, the argument of you cube roots of $-27$ and as given in your answer, has three values that it can take. [Note that for all three $\alpha$s, the value for the modulus of the roots $($or $q)$ is always the same ($-3$). This is where your problem came about. (You didd not consider other possible roots)]

So now, we have found all values that a complex number, say $u$ can take where $$u^3=-27$$ And your question asks you to find the solution to the equation $$(z - 6 + \text i)^3 = -27$$

Notice the similarity? They are in similar forms right? So it would be fair to say that

$$~~~~~~~~~~~~~~u=(z - 6 + \text i)$$ $$\implies z=u+6-\text i \tag{3}$$ From this you easily get 3 values for $z$ (using the three solutions for $u$ that we got earlier on)

Note: During the last step, you need to change solutions for $u$ from their Polar form to the usual $a+b\text i$ format. The reason should be apparent from $(3)$
Hope it helps!

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The other approach to this problem could be to rewrite equation into form :

$z^3-(18-3i)z^2+(105-36i)z-(171-107i)=0$

and then to use general formula for the roots of the cubic equation .

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Since one solution $z_0$ is already known I'd rather divide that expression by $z-z_0$ and end up with a quadratic equation than use that formula. – user20266 Apr 1 '12 at 10:52
@Thomas Good point.... – pedja Apr 1 '12 at 10:53