AC circuit with two sources
are they in direct current?


No.  A real valued phasor is still a phasor with a phase of zero.  The time dependence has been 'hidden' but one must always keep in mind that the phasor value is the amplitude and phase of a sinusoidal function of time.

If, in fact, the frequency of the sinusoid were zero, all capacitors would be have infinite impedance and all inductors would have zero impedance.

The fact that this isn't the case in the circuit given means that the actual voltages and currents are sinusoids of non-zero frequency.

This circuit is easily solved by superposition which allows the solution to be written by inspection.

For example, the (phasor) voltage across the current source is, by superposition:

$$1V \cdot \frac{j1}{j1 + (-j1)||1} + 2A \cdot j1||(-j1)||1 \Omega $$

The first term is with the current source off and is an application of voltage division.

The second term is with the voltage source off and is just the current multiplied by the equivalent impedance seen by the source.

Since the frequency and/or capacitance and inductance values are not given, one cannot convert the phasor solutions to time dependent functions.  One might as well consider the (angular) frequency to be \$\omega = 1\$ and the capacitance and inductance to be \$C = 1F\$ and  \$L = 1H\$ respectively.

For a different frequency, the values of the capacitance and inductance would scale appropriately such that impedances are unchanged.