How to solve the equation z^4=ibarz ? Where z is a complex number.

Mar 5, 2017

$S = \left\{0 , {e}^{\frac{i \pi}{10}}\right\}$

Explanation:

Let's think of these complex numbers in exponential forms!

$z$ is a complex number of phase, or angle $\theta$, and its conjugate has the angle $- \theta$, we can see that by algebra if we use certain trig properties (the cosine is even and the sine is odd):

$z = r {e}^{i \theta} = \cos \left(\theta\right) + i \cdot \sin \left(\theta\right) = a + b i$
$\overline{z} = r {e}^{- i \theta} = \cos \left(- \theta\right) + i \sin \left(- \theta\right) = \cos \left(\theta\right) - i \sin \left(\theta\right) = a - b i$

So, rewriting this in terms of angles, we have:

$r {e}^{i \cdot 4 \theta} = i \cdot r {e}^{- i \theta}$

Remembering that $i$ also has an exponential form, ${e}^{\frac{\pi}{2} \cdot i}$

$r {e}^{i \cdot 4 \theta} = r {e}^{i \left(\frac{\pi}{2} - \theta\right)}$
$i \cdot 4 \theta = i \left(\frac{\pi}{2} - \theta\right)$
$4 \theta = \frac{\pi}{2} - \theta$
$5 \theta = \frac{\pi}{2}$
$\theta = \frac{\pi}{10}$

Now to solve for the radius, we know that:

${z}^{4} = i \overline{z}$

By taking the absolute value of both sides we have

$| {z}^{4} | = | i \overline{z} |$

Which we can split

$| z {|}^{4} = | i | | \overline{z} |$

But $| z | = | \overline{z} |$ and it's defined as the radius we want to discover, so

${r}^{4} = | i | \cdot r$

$| i | = 1$, so simplifying we have

${r}^{4} = r$

There are only two real numbers (remember that the absolute value / radius is always real) that can satisfy this equation

$r = 1$ or $r = 0$

Mar 5, 2017

$z = 0 , i$
$z = \pm \frac{1}{2} \sqrt{\frac{1}{2} \left(5 + \sqrt{5}\right)} + \frac{1}{4} \left(\sqrt{5} - 1\right) i$
$z = \pm \frac{1}{2} \sqrt{\frac{1}{2} \left(5 - \sqrt{5}\right)} - \frac{1}{4} \left(\sqrt{5} + 1\right) i$

Explanation:

We have:

${z}^{4} = i \overline{z}$,

Where $\overline{z}$ is the complex conjugate of $z$

If we use the exponential form of a complex number then we can represent the solution by;

$z = r {e}^{i \theta} \setminus \setminus \setminus \left(= r \cos \theta + i r \sin \theta\right)$

Then the complex conjugate will be:

$z = r {e}^{- i \theta}$

Whenever dealing with complex equation such as this it is essential to remember that the complex exponential has a period of $2 \pi$, and thus we can modify the conjugate, and substitute into the equation to solve to get;

${\left\{r {e}^{i \theta}\right\}}^{4} = i \left(r {e}^{- i \theta}\right)$
${\left\{r {e}^{i \theta}\right\}}^{4} = i \left(r {e}^{i \left(- \theta + 2 n \pi\right)}\right)$ where $n \in \mathbb{N}$

(Equally we could have incorporated this periodicity into the RHS exponent, it really does not matter providing it is incorporated somewhere). Replacing $i$ by its complex exponential form $i = {e}^{i \frac{\pi}{2}}$ we get:

${\left\{r {e}^{i \theta}\right\}}^{4} = {e}^{i \frac{\pi}{2}} \left(r {e}^{i \left(- \theta + 2 n \pi\right)}\right)$
$\therefore {r}^{4} {e}^{4 i \theta} = r {e}^{i \left(\frac{\pi}{2} - \theta + 2 n \pi\right)}$

Equating real and imaginary terms (or the moduli and arguments) we have:

$R e \setminus : \setminus \setminus {r}^{4} = r$
$I m : \setminus \setminus 4 \theta = \frac{\pi}{2} - \theta + 2 n \pi$

From the first equation we get:

${r}^{4} - r = 0$
$\therefore r \left({r}^{3} - 1\right) = 0$
$\therefore r = 0 , 1$

From the second equation we get:

$4 \theta = \frac{\pi}{2} - \theta + 2 n \pi$
$\therefore 5 \theta = \frac{\pi}{2} + 4 n \frac{\pi}{2}$
$\therefore \theta = \left(4 n + 1\right) \frac{\pi}{10}$

So now we can "assemble" the solutions:

$r = 0 \implies z = 0$

$r = 1 \implies z = {e}^{i \theta}$

And By putting $n = 0 , 1 , 2 , 3 , \ldots$ we get:

$\theta = \frac{\pi}{10} , \frac{\pi}{2} , \frac{9 \pi}{10} , \frac{13 \pi}{10} , \frac{17 \pi}{10} , \ldots$

And with these values of $\theta$, and using $z = \cos \theta + i \sin \theta$ we get:

$n = 0 : \implies z = \frac{1}{2} \sqrt{\frac{1}{2} \left(5 + \sqrt{5}\right)} + \frac{1}{4} \left(\sqrt{5} - 1\right) i$
$n = 1 : \implies z = 0 + 1 i$
$n = 2 : \implies z = - \frac{1}{2} \sqrt{\frac{1}{2} \left(5 + \sqrt{5}\right)} + \frac{1}{4} \left(\sqrt{5} - 1\right) i$
$n = 3 : \implies z = - \frac{1}{2} \sqrt{\frac{1}{2} \left(5 - \sqrt{5}\right)} - \frac{1}{4} \left(\sqrt{5} + 1\right) i$
$n = 4 : \implies z = \frac{1}{2} \sqrt{\frac{1}{2} \left(5 - \sqrt{5}\right)} - \frac{1}{4} \left(\sqrt{5} + 1\right) i$

After which the pattern repeats. Hence there are six solutions:

$z = 0 , i$
$z = \pm \frac{1}{2} \sqrt{\frac{1}{2} \left(5 + \sqrt{5}\right)} + \frac{1}{4} \left(\sqrt{5} - 1\right) i$
$z = \pm \frac{1}{2} \sqrt{\frac{1}{2} \left(5 - \sqrt{5}\right)} - \frac{1}{4} \left(\sqrt{5} + 1\right) i$
We can see these solutions on the Argand diagram: