Another A.C circuit question =/

Not sure if the answer to this will be as easy as the last one

i) The impedance of teh capactiro expressed in polar form.

A) Cz = V/i
= 79.62 < -90 / 2.5 < 0
= 31.848 < -90

ii) The capacitance of the capacitor
A) C = 1/(2 pi x f x Xc)
= 1 / (2pi x 100 x 31.848)
= 50uF

(iii)The Voltage across the coil in polar form

So...

XL = 2 Pi x f x L

= 2pi x 100 x 47mH

= 29.53

To work out the voltage drop across it i use V = I * R (or V = I * Z)
= 2.5 < 0 * 29.52
Vcoil = 73.38 < 0

But in rectangle form it would be 73.38+j0 which isnt right :S

What have i done wrong

2.5 x 29.52 = 73.8 not 73.38........... And you evaluated it as 29.53 not 29.52. Sloppy. 73.825.
You haven't given the Whole Question, just introduced loads of numbers from nowhere.
Are you sure that the Answer they were looking for IS Vcoil ??
Vcoil = 73.825 < +90 !!!!! if i is at angle Zero.
Hence in rectangular form it would be 0 + j73.825.
I have not corrected any rounding errors which may have accumulated.
Moreover, I had to make LOTS of assumptions regarding the QUESTION!!!!.
If this refers to some earlier question, then I haven't seen it.