Polar complex form and exponent question I don’t understand how exponents impact the angle $\phi$.
My understanding:
$\quad$ Let’s say we have a number: 
$$(\cos{(3\pi/2)}+i\sin{(3\pi/2)})^{3} = (\cos{(\frac{3\pi*3}{2}+2k\pi)}+i\sin{(\frac{3\pi*3}{2}+2k\pi)})$$
I understand this, we multiply $\phi$ with exponent and $+2k\pi \,$is the period of the function. And of course there is $|z|^{3}$ multiplied with the whole expression.
$\quad$I don’t understand this:
$$(\cos{(3\pi/2)}+i\sin{(3\pi/2)})^{1/4}= (\cos{(\frac{3\pi+2k\pi}{2*4}}+i\sin{(\frac{3\pi+2k\pi}{2*4})})$$
Why is $2k\pi$ in the numerator if the exponent is less than one. 
What to do if the exponent is $3/4$ or $7/4$ or some other fraction?
Maybe I written wrong in my notes and $+2k\pi$ is always out of the fraction
 A: Hint: start with
$$
\cos{(3\pi/2)}+i\sin{(3\pi/2)}=\cos{(3\pi/2+2\pi k)}+i\sin{(3\pi/2+2\pi k)}
$$
and then apply the power of $1/4$ to both sides.
A: it should be $$\frac 14(\frac{3\pi}{2}+2k\pi)$$
A: Observe that
$$\cos(\phi)+i\sin(\phi)=\cos(\phi+2k\pi)+i\sin(\phi+2k\pi).$$
The angle $\phi$ is indeterminate to a multiple of $2\pi$.
This is why you must account for that when multiplying by a non-integer fraction - i.e. taking a rational power. 
$$\frac\phi4\ne\frac{\pi+2\pi}4\ne\frac{\pi+4\pi}4\ne\frac{\pi+6\pi}4$$
but
$$\frac\phi4=\frac{\pi+8\pi}4-2\pi,$$
$$\frac{\phi+2\pi}4=\frac{\pi+10\pi}4-2\pi,$$ and so on.
Multiplying by an integer is harmless, a multiple of $2k\pi$ remains a multiple of $2k\pi$. And multiplying by an irrational is problematic (see why ?).
A: To simplify notation let us replace $3\pi/2$ by $t$ and use the formula $\cos(t) + i \sin(t) = e^{it}$. The function $t \mapsto e^{it}$ has period $2\pi$.
Thus you get
$$e^{it} = e^{i(t + 2 k\pi)}$$
and
$$(e^{it})^n = e^{int} = e^{i(nt + 2 k\pi)} .$$
You see that the summand $2k\pi$ is irrelavant here. However, it you want to determine the $n$-th root of $e^{it}$, you must be aware that you have $n$ such roots. The "naive" solution is
$$(e^{it})^{\frac{1}{n}} = e^{i\frac{t}{n}}, $$
but of course also the values
$$e^{i\frac{t + 2k\pi}{n}}$$
will be solutions. Note that these values only depend only $k \mod n$, that is, you get distinct roots for $k =0,\ldots, n-1$, and you cannot omit $2k\pi$ from the general solution.
Thus your equation has to be corrected to
$$(\cos{(3\pi/2)}+i\sin{(3\pi/2)})^{1/4}= \cos(\frac{3\pi}{2*4}+ \frac{2k\pi}{4})+i\sin(\frac{3\pi}{2*4}+ \frac{2k\pi}{4}) .$$
