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Chapter 6: Problem 53

Use DeMoivre's Theorem to find the indicated power of the complex number. Write answers in rectangular form. $$\left[4\left(\cos 15^{\circ}+i \sin 15^{\circ}\right)\right]^{3}$$

Short Answer

Expert verified
So, \(\left[4\left(\cos 15^{\circ}+i \sin 15^{\circ}\right)\right]^{3} = 32\sqrt{2} + 32\sqrt{2}i\).

Step by step solution

01

Analyzing the given complex number

The given complex number is in the trigonometric(polar) form: \(4(\cos 15^{\circ}+i \sin 15^{\circ})\). Here, the modulus is 4 and the argument is \(15^{\circ}\). We are asked to find the cube of this complex number.
02

Apply DeMoivre鈥檚 theorem

According to DeMoivre鈥檚 theorem, \((r(\cos 胃 +i\sin 胃))^{n} = r^{n}(\cos n胃 +i\sin n胃)\). Here, \(r=4\), \(胃 = 15^{\circ}\) and \(n=3\). Plugging these values in, we get: \((4)^{3}(\cos (3\cdot15^{\circ}) + i\sin (3\cdot15^{\circ})) = 64\cos 45^{\circ} + 64i \sin 45^{\circ}\).
03

Convert back to rectangular form

In rectangular form, a complex number is written as \(a+bi\). Use the known values of \(\cos 45^{\circ}\) and \(\sin 45^{\circ}\) which are both \(\sqrt{2}/2\), we can calculate \(a\) and \(b\) as follows: \(a = 64 \cdot \sqrt{2}/2 = 32\sqrt{2}\) and \(b = 64 \cdot \sqrt{2}/2 = 32\sqrt{2}\). So the rectangular form of given complex number's cube is \(32\sqrt{2} + 32\sqrt{2}i\)

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