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Chapter 6: Problem 63

Use DeMoivre's Theorem to find the indicated power of the complex number. Write answers in rectangular form. $$(\sqrt{3}-i)^{6}$$

Short Answer

Expert verified
So \( (\sqrt{3} - i)^6 = 64 \)

Step by step solution

01

Convert complex number to polar form

Given \( (\sqrt{3}-i) \), we convert this to polar form: \( r(cos(\Theta) + isin(\Theta)) \) where \( r = \sqrt{Re^2 + Im^2}\) and \( \Theta = arctan(\frac{Im}{Re})\). So \( r = 2\) and \( \Theta = 330^{\circ} \) because \( \sqrt{3} \) is in the fourth quadrant.
02

Apply DeMoivre's Theorem

Next we raise the complex number to the indicated power using DeMoivre's Theorem, which states: \( (r(cos(\Theta) + isin(\Theta)))^n = r^n(cos(n\Theta) + isin(n\Theta)) \). Therefore: \( 2^6(cos(6*330^{\circ}) + isin(6*330^{\circ}))\).
03

Convert back to rectangular form

To convert back to rectangular form, we apply the conversion formula which is \( rcos(\Theta) + r isin(\Theta) = r(\cos(\Theta) + i \sin(\Theta)) \). Therefore, we have \( 64(cos(0) + isin(0))=64(1+i0)=64+i0 \).

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