How calculate impedance of series resonance LCR circuit?

not sure

using the j operator, and w to represent angular frequency,
Z = R + jwL - i/jwC
When |Xc| and |Xl|, Z = R
To draw a phase diagram, plot R as + horizontal, Xl  as + vertical, and Xc as - vertical.  Z is the resultant.

You can just add impedance in series- R value for resistors, jwL for inductors,1/(jwC) for capacitors, and don't forget phase. (Where w is frequency, C is capacitance, L is inductance, R is resistance, and j is the imaginary number sqrt(-1)

A phase diagram has frequency on the x-axis and phase on the y-axis; you can use a Bode plot

Helmut has it correct.   This occurs when wL = 1/wC.  The frequency in radians/sec then becomes w = (LC)^-1/2  ... in other words, the square root of 1/LC.  At resonance, these terms cancel out and you are left with a purely real impedance = resistance of the circuit which has a phase angle of 0.

The phase angle at resonance can be plotted on the complex plane where the x axis represents real numbers (R) and the y axis represents imaginary numbers (Xc and Xl).  Since Xc and Xl are equal with opposite phase (+j and -j = +90 degrees and -90 degrees), they cancel out leaving only R on the plot with a phase angle of 0 degrees since R remains on the x axis and the phase angle is found as the angle from the postive x axis to the plotted point on the complex plane.  

Bode plots do not really apply since we do not know values of L and C and the relationship makes a big difference in the full plot  of angle or magnitude versus frequency and the question was for resonance which we know will be 0 degrees since it is purely real impedance at resonance.