The circuit is here:
http://i36.tinypic.com/2ro59ty.png

 

The two impedances are in series so you draw 10 ohms along the positive x-axis which is the real axis and you draw 15 ohms alond the vertical axis which is the imaginary or jω-axis. The impedance is the vector sum of 10 + j15

|Z] = SQRT(10^2 + 15^2) = SQRT(325) ≈ 18
arg(Z) = tan^-1(15/10) ≈ 56.3°

Was that what you were looking for?

 

Yes! that helped me a lot. I had that down on my paper, but was unsure if I was correct. Thank you!

I have another question also.
Circuit:
http://i33.tinypic.com/331ese9.png

How do i find the ZT, Degrees, VR, VC, and VL?

 

XL is ploted along the positive jω-axis and Xc is plotted along the negative jω-axis so:

170 - 90 = 80

so the capacitor in series with the inductor looks like an inductor with a smaller reactance namely 80 ohms. From there it is just like the previous problem