Answer
a.) $I(1000t) = 0.013e^{j\frac{\pi}{6}} $ A
b.) $-2000j$ Ohms
c.) $+26e^{-j\frac{\pi}{3}}$V (top terminal is + and bottom is - on capacitor)
Work Step by Step
start by using euler identity to convert to phasor notation
a.) $I(1000t) = 0.013e^{j\frac{\pi}{6}}$
b.) $Z = \frac{1}{j*w*C} = -\frac{1}{1000*(0.5*10^{-6})*j} = -2000j$
Now recall that $Z = V/I = \frac{1}{jwC}$ (For a capacitor)
Multiply both sides by I(wt) to solve for V(wt)
$(2000e^{-j\frac{\pi}{2}})(0.013e^{j\frac{\pi}{6}}) = V(wt) $
$26e^{-j\frac{\pi}{3}} = V(wt)$
Polarity is determined by the passive sign convention. With current flowing into the positive terminal of the capacitor.
