It seems you have understood it correctly, but you may also have done it unnecessarily complicated. Because, using only algebra, you can construct a meaningful, if not intuitive, at least more inutitive way of representing complex exponentiation:
Let $z = a+bi$ and $w = c+di$. Note: $z = |z|*e^{Arg(z)*i}$
$$z^w = (a+bi)^{c+di}$$$$ = (a+bi)^c*(a+bi)^{di}$$$$ = |z|^c*e^{iArg(z)c}*|z|^{di}*(e^{Arg(z)*i})^{di}$$$$= |z|^c*e^{-Arg(z)d}*e^{ln(|z|)di}*e^{iArg(z)c}$$$$ = \frac{|z|^c}{e^{Arg(z)d}}e^{(Arg(z)c+ln(|z|)d)i}$$
$|z| = \sqrt{Re(z)^2+Im(z)^2}$ ➝ The length of z
$Arg(z) = atan2(Im(z), Re(z))$ ➝ The angle of z
$e^{i\theta} = cos(\theta)+isin(\theta)$ ➝ Euler's formula
Using this formula, you only need to know how to do exponentiation with real numbers, finding the angle of a complex number, finding ln(x), finding |x|, euler's formula, multiplication, division, addition and sometimes subtraction.
Some interesting observations I've made:
- When taking any complex or real number to the power of $i$, the angle and length of the number seem to swap: $z = a+bi, w = 0+1i$
$$z^w = \frac{|z|^0}{e^{Arg(z)*1}}e^{(Arg(z)*0+ln(|z|*1)i} = \frac{1}{e^{Arg(z)}}e^{ln(|z|)i}$$
When taking $z^i$, $1$ divided by $e$ to the power of the angle of $z$, is now the length, and the natural logarithm of the length is now the angle.
- Doing complex exponentiation with a real exponent is essentially a mix of stretching and rotating the base.
- Doing complex exponentiation with an imaginary exponent basically just swaps the length and the angle.
- Doing complex exponentiation with a complex exponent is a mix of the two things mentioned above.
Complex roots
Finding complex roots is very similar to taking complex exponentials, since, as we know: $\sqrt[b]{a} = a^{\frac{1}{b}}$
Having used some basic algebra, I've come up with a similar formula for the wth root of z where $z,w ∈ ℂ$:
$$\sqrt[w]{z} = z^{\frac{1}{w}}$$
$$= exp\left(\frac{Arg(z)d+ln(|z|)c}{c^2+d^2}+\frac{Arg(z)c-ln(|z|)d}{c^2+d^2}i\right)$$
Best Answer
Note that:
$$z = re^{i\phi}, z^5 = r^5 e^{i5\phi}~\text{and}~\bar{z} = re^{-i\phi},$$
where $r \geq 0$. Therefore:
$$z^5 = \bar{z} \Rightarrow \begin{cases} r^5 = r\\ 5\phi + 2k\pi= -\phi + 2h\pi \end{cases},$$
where $k, h \in \mathbb{Z}.$
The previous system can be rewritten as: $$ \begin{cases} r(r^4-1) = 0\\ 6\phi = 2s\pi \end{cases},$$
where $s = k-h \in \mathbb{Z}.$
The first equation has $3$ distinct roots: $r=-1$, $r=0$ and $r=1.$ Of course, $r=-1$ should be discarded. This means that for $r=0$, $z = 0$ is a solution, which obviously does not depend on the phase $\phi.$ Moreover, for $r=1$, the phase is important. Solving the second equation, we get:
$$\phi = \frac{s\pi}{3},$$
and hence $z = e^{\frac{is\pi}{3}}$ for $s \in \mathbb{Z}$, together with $z=0$ represent the solution of the equation.