Four quadrant description
Lets assume your phasors rotate in a technically positive direction(counter-clock). So the phasors will be a representation of passed time in regard to the \$x\$ axis. Lets look at it through the formula:
$$ \underline{I}=Ie^{j\varphi}$$
is actually a representation of:
$$ I(t)=I\sqrt{2}[\cos(\omega t +\varphi)+j\sin(\omega t + \varphi)] $$
A deeper theoretical elaboration can be found at this link.

You usually represent current and voltage in a phasor diagram. You can represent reactive and active power too, but that won't be called a phasor diagram in strict interpretation.
It is usual to put the voltage \$\underline{U}\$ as reference on the \$x\$-axis. Doing this we simplify the representation without loss of generality. So our basic system will be:
$$\underline{U}=Ue^{j0^o}\\
\underline{I}=Ie^{j\varphi}$$
The phase angle will be put next to the current so that we have a framework for describing phase differences.

There are three basic kinds of impedances: Restive(\$\mathcal{Z}_R\$), Capacitive(\$\mathcal{Z}_C\$) and inductive(\$\mathcal{Z}_L\$). Combinations are possible of course. They are represented as:
$$\mathcal{Z}_R=R=\mathfrak{Re}\{R\}\\
\mathcal{Z}_C=X_Ce^{-j90^o}=\frac{1}{j\omega C}\\
\mathcal{Z}_L=X_Le^{j90^o}=j\omega L$$
The generalised impedance can be represented as:
$$\mathcal{Z}=R+jX=|\mathcal{Z}|e^{-j\varphi}$$
You should now ask why does the argument of \$\mathcal{Z}\$ has a minus sign. The reason behind this is:
$$\underline{I}=\frac{Ue^{j0^o}}{|\mathcal{Z}|e^{-j\varphi}}\\
\underline{I}=Ie^{j\varphi}$$

Lets now describe the terminology that is used for this system:


Inductive: Means that your load is of dominantly inductive character. The current will lag behind the voltage so the phase angle will be \$-90^o&lt;\varphi&lt;0 \$. Differential equation for inductor: \$u_L(t)=L\frac{\mathbb{d}i_L}{\mathbb{d}t}\$.
Capacitive: Means that your load is of dominantly capcitive character. The voltage will lag behind the current so the phase angle will be \$0^o&lt;\varphi&lt;90^o \$. Differential equation for capacitor: \$i_c(t)=C\frac{\mathbb{d}u_c}{\mathbb{d}t}\$.
Resistive: Only the real part exists. \$\varphi=0\$
Angles that are in \$(90^o,270^0)\$ represent only that the flow direction of current is inverse, meaning \$\underline{I}=-\underline{I}_{negative}\$. Here is a graph for a capacitive resistive load, the amplitudes of \$U,I,\mathcal{Z}\$ are arbitary as the units are different.








simulate this circuit &ndash; Schematic created using CircuitLab