How do I find the power of (3-i)^12 using trigonometric form?

[hellbike]

how to find number $$(3-i)^{12}$$ ?

i know theorem for powers of complex numbers, but i have to know argument of trygonometrical form.

and its $$cos x = 3/4$$ and $$sin x = -1/4$$

and i don't know what to do with that.

 

[g_edgar]

First write $3 - i$ in the form $r (\cos\theta + i \sin \theta)$ . Then $(3 - i)^{12}$ will be $r^{12} (\cos(12 \theta) + i \sin (12 \theta))$