Help with complex numbers

Question

Algebraic form of the $z^3 = (3 + i)^6$. Can someone help me to solve this? My answer is $z = 8 + 6i$, but I'm not sure that this is true.

Answer 1

Let

$$\left(3 + i\right)^2 = 8 + 6i = a \tag{1}\label{eq1}$$

so

$$z^3 = \left(3 + i\right)^6 = \left[\left(3 + i\right)^2\right]^3 \tag{2}\label{eq2}$$

becomes, after moving the right side to the left & factoring, that

$$z^3 - a^3 = \left(z - a\right) \left(z^2 + za + a^2\right) = 0 \tag{3}\label{eq3}$$

Thus, either $z - a = 0$ or $z^2 + za + a^2 = 0$. For the first one $z = a$ becomes, from \eqref{eq1}, that $z = 8 + 6i$, as you have already determined. For the second one, you can just use the quadratic formula to get the other $2$ roots as

$$z = \cfrac{-a \pm \sqrt{-3a^2}}{2} = \cfrac{-a \pm \sqrt{-3}a}{2} \tag{4}\label{eq4}$$

Note this becomes

$$z = \left(-\cfrac{1}{2} \pm \cfrac{\sqrt{3}}{2}i\right)a \tag{5}\label{eq5}$$

as orion has previously stated in their answer.

Answer 2

You found one solution. The other two can be found by rotating the known solution in the complex plane by $\pm 120^\circ$ ($n$-th roots always form a regular $n$-sided polygon). For rotation, you can multiply by $\cos\frac{2\pi}{3}\pm i\sin\frac{2\pi}{3}=-\frac12\pm i\frac{\sqrt{3}}{2}$.

Answer 3

Whenever you have $z^3 = k$ there will be $3$ solutions.

If $z^3 = (3+i)^6 = ((3+i)^2)^3$ then one solution is obviously $z = (3+i)^2 = 9 + 6i + i^2 = 9+6i -1 = 8+6i$.

But that is only one solution.

However $w^3 = 1$ has three solutions.

If we let $w$ be one of the solutions and we let $z = w(8+6i) = w(3+i)^2$ then $z^3 = w^3((3+i)^2)^3 = 1*(3+i)^6 = (3+i)^6$.

So if the three solutions to $w^3=1$ are $1, b,c$ then the three solutions to $z^3 = (3+i)^6$ are $8 + 6i, b(8+6i), c(8+6i)$.

So we nee to solve $w^3 = 1$.

Let $w = m+ni$ (where $m,n$ are real) then $(m+ni)^3 = m^3 + 3m^2ni + 3mn^2i^2 + n^3i^3=$

$m^3 + 3m^2ni -3mn^2 - n^3i = $

$(m^3 - 3mn^2) + (3m^2n-n^3)i = $

$1 + 0i$

So $m^3 -3mn^2 =1$ and $3m^2n -n^3=0$.

$3m^2n -n^3= n(3m^2-n^2) = n(m\sqrt 3 -n)(m\sqrt 3 + n) = 0$ so

either:

1) $n = 0$ and $m^3 = 1$ so $m = 1$ and $w = 1 + 0i = 1$.

2) $m\sqrt 3 - n = 0$ and $m\sqrt 3 = n$ so $m^3 -3mn^2 = 1$ and so

$m^3 - 3m(m\sqrt 3)^2 = m^3 - 3m(3m^2) =m^3 -9m^3 = -8m^3 =1$ so $m = -\frac 12$ and $w = -\frac 12 -\frac {\sqrt 3}2 i$.

3) $m\sqrt 3+ n = 0$ so $n = -m\sqrt 3$ and so

$m^3 - 3m(-m\sqrt 3)^2=.... = 1$ so $m = -\frac 12$ and $w = -\frac 12 + \frac {\sqrt 3}2i$.

So the three solutions to $z^3 =(3+i)^6$ are:

1)$8+6i$

2)$(-\frac 12 -\frac {\sqrt 3}2 i)(8 + 6i) = -4 -4\sqrt 3i - 3i -3\sqrt 3 i^2 = (-4+3\sqrt 3) -(4\sqrt 3-3)i$.

and 3) $(-\frac 12 +\frac {\sqrt 3}2 i)(8 + 6i) = -4 +4\sqrt 3i - 3i +3\sqrt 3 i^2 = (-4-3\sqrt 3) +(4\sqrt 3-3)i$

....

However once you learn polar notation that $z = m + ni = r(\cos \theta + i\sin \theta) := re^{i\theta}$ where $r = \sqrt{m^2 + n^2}$ and $\theta$ is the solutions to $\cos \theta = \frac mr$ and $\sin \theta = \frac nr$, once you learn that these problems become MUCH easier.