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Chapter 6: Problem 64

Use DeMoivre's Theorem to find the indicated power of the complex number. Write answers in rectangular form. $$(\sqrt{2}-i)^{4}$$

Short Answer

Expert verified
The answer is -9.

Step by step solution

01

Convert to Polar Form

The polar form of a complex number \(z = a + bi\) is \(r(\cos{\theta} + i\sin{\theta})\), where \(r = \sqrt{a^2 + b^2}\) and \(\theta = \arctan{\frac{b}{a}}\). For the given complex number \(\sqrt{2}-i\), \(a = \sqrt{2}\) and \(b = -1\). So, \(r = \sqrt{(\sqrt{2})^2 + (-1)^2} = \sqrt{2 + 1} = \sqrt{3}\), and \(\theta = \arctan{\frac{-1}{\sqrt{2}}} = -\frac{\pi}{4}\). Therefore, \(\sqrt{2}-i = \sqrt{3}(\cos{-\frac{\pi}{4}} + i\sin{-\frac{\pi}{4}})\).
02

Use DeMoivre's Theorem

DeMoivre's Theorem states that \([r(\cos{\theta} + i\sin{\theta})]^n = r^n(\cos{n\theta} + i\sin{n\theta})\). Applying this rule, \((\sqrt{3}(\cos{-\frac{\pi}{4}} + i\sin{-\frac{\pi}{4}}))^4 = (\sqrt{3})^4(\cos{-\frac{4\pi}{4}} + i\sin{-\frac{4\pi}{4}}) = 9(\cos{-\pi} + i\sin{-\pi})\).
03

Convert Back to Rectangular Form

The rectangular form of a complex number is \(r(\cos{\theta} + i\sin{\theta}) = r\cos{\theta} + ir\sin{\theta}\). Applying this, we get \(9(\cos{-\pi} + i\sin{-\pi}) = 9\cos{-\pi} + 9i\sin{-\pi} = -9\).

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