91Ó°ÊÓ

To convert a complex number in rectangular form, \(a + bi\), to polar form, we need to find the modulus (r) and the argument (θ). We have: Modulus: \(r = \sqrt{a^2 + b^2}\) Argument: \(\theta = \arctan \left(\frac{b}{a}\right)\) For our complex number, \(a = -3\) and \(b = 3\sqrt{3}\). Let's compute the modulus and argument: Modulus: \(r = \sqrt{(-3)^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = 6\) Argument: \(\theta = \arctan \left(\frac{3\sqrt{3}}{-3}\right) = \arctan(-\sqrt {3}) = \frac{4\pi}{3}\) Now, we can write the complex number in polar form as: \(6(\cos{\frac{4\pi}{3}} + i \sin{\frac{4\pi}{3}})\).

According to De Moivre's Theorem, if we have a complex number in polar form raised to a power, we can calculate it as follows: \((r(\cos{\theta} + i \sin{\theta}))^n = r^n (\cos{n\theta} + i \sin{n\theta})\) For our complex number and the given power 555, we compute: \((6(\cos{\frac{4\pi}{3}} + i \sin{\frac{4\pi}{3}}))^{555}= 6^{555}(\cos{\frac{2220\pi}{3}} + i \sin{\frac{2220\pi}{3}})\)

Now we need to convert the result back to rectangular form. Recall the trigonometric identities: \(r\cos{\theta} = a\) \(r\sin{\theta} = b\) From our previous step, we have: \(r = 6^{555}\) \(\theta = \frac{2220\pi}{3}\) We can simplify the polar form by dividing the angle by \(2\pi\): \(\frac{2220\pi}{3} = n \cdot 2\pi\) \(n = \frac{1110}{3} = 370\) Since the cosine and sine functions are periodic, we can find the sine and cosine of the simplified angle \(\frac{4\pi}{3}\). \(a = 6^{555}\cos{\frac{4\pi}{3}} = 6^{555}\cdot \left(-\frac{1}{2}\right)\) \(b = 6^{555}\sin{\frac{4\pi}{3}} = 6^{555}\cdot\left(-\frac{\sqrt{3}}{2}\right)\) Finally, the answer in rectangular form is: \((-3 + 3 \sqrt{3} i)^{555} = \left(-\frac{1}{2} \cdot 6^{555}\right) + \left(-\frac{\sqrt{3}}{2} \cdot 6^{555}\right)i\)