so we have this circuit :

Considering \( V_a \) to be the origin of phase-voltage so \( V_{ab} \)

We have : \( V_{ab} = V_a \sqrt{3} (30°) \\

V_{bc} = V_b \sqrt{3} (-90°) \\

V_{ca} = V_c \sqrt{3} (150°)\)

and we have \( I_a = \frac{V_a}{|Z|} (-arg(Z)) \\

I_b = \frac{V_b}{|Z|} (-arg(Z)) \\

I_c = \frac{V_c}{|Z|} (-arg(Z)) \)

Thus:

\( I_a = \frac{V_{ab}}{\sqrt{3}|Z|} (-30-arg(Z)) \\

I_b = \frac{V_{bc}}{|Z|} (90-arg(Z)) \\

I_c = \frac{V_{ca}}{|Z|} (-150-arg(Z)) \)

with \( V_{ab} = 230 V = V_{bc} = V_{ca} \)