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

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SUMMARY

The discussion focuses on calculating the power of the complex number (3 - i) raised to the 12th power using trigonometric form. The argument is established with cos(θ) = 3/4 and sin(θ) = -1/4, leading to the conversion of (3 - i) into the form r( cos(θ) + i sin(θ)). The final result is derived using the formula (3 - i)^{12} = r^{12} (cos(12θ) + i sin(12θ)).

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  • Familiarity with trigonometric identities
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  • Basic skills in using De Moivre's Theorem
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hellbike
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how to find number [tex](3-i)^{12}[/tex] ?

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

and its [tex]cos x = 3/4[/tex] and [tex]sin x = -1/4[/tex]

and i don't know what to do with that.
 
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First write [itex]3 - i[/itex] in the form [itex]r (\cos\theta + i \sin \theta)[/itex] . Then [itex](3 - i)^{12}[/itex] will be [itex]r^{12} (\cos(12 \theta) + i \sin (12 \theta))[/itex]
 

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