Why does a pure resistance has a an imaginary current and voltage?

Because the circuit is not purely resistive. The imaginary part is added by the inductor in this case.
The same current must go through resistor and inductor. See how this in influences your calculations and the real and imaginary parts.

I'll also move this thread to where it belongs: homework.

 

What is the equation for a voltage across a resistor?

When you look at the voltage across and the current through the resistor only, you will see that both are perfectly in phase. Since the current has an imaginary part, so does the voltage.
Real and imaginary make sense only in combination with a reference frame that establishes a 0 ° phasor (or a purely real signal). Then you can determine the phase of other signals (voltages, currents) with reference to that 0 ° signal. When you look at a single signal, isolated from the others in the system/circuit, phase makes not much sense (if any at all).
In your example the reference signal with phase 0 ° would be the driving voltage source. Only when looking at the voltage with respect to this 0 ° reference will you "see" the imaginary part of the voltage across the resistor. You will not "see" the imaginary part when you look at the voltage and current across and through the resistor only.