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

hellbike
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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.
 
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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))
 

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