Question

assignment 6: problem 14
(1 point)

consider the equation ( z^{17}=(1+sqrt{3} i) ). find the value of ( z ) which satisfies this equation and which has the second smallest positive argument ( \theta, 0<\theta<2 pi ). express your answer as ( z=r e^{i \theta} ) where
( r=quad ) and ( \theta=)

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Explanation:

Step1: Convert \(1+\sqrt{3}i\) to polar form

For a complex number \(z = a+bi\), the modulus \(r=\sqrt{a^{2}+b^{2}}\) and the argument \(\theta=\tan^{- 1}(\frac{b}{a})\) (adjusted for the quadrant).
For \(z = 1+\sqrt{3}i\), \(r=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{1 + 3}=2\), \(\theta=\tan^{-1}(\frac{\sqrt{3}}{1})=\frac{\pi}{3}\). So, \(1+\sqrt{3}i = 2e^{i\frac{\pi}{3}}\)

Step2: Solve \(z^{17}=2e^{i\frac{\pi}{3}}\)

If \(z = re^{i\theta}\), then \((re^{i\theta})^{17}=r^{17}e^{i17\theta}\).
We have \(r^{17}e^{i17\theta}=2e^{i(\frac{\pi}{3}+2k\pi)}\), \(k\in\mathbb{Z}\).
From \(r^{17}=2\), we get \(r = 2^{\frac{1}{17}}\)
From \(17\theta=\frac{\pi}{3}+2k\pi\), we get \(\theta=\frac{\pi}{51}+\frac{2k\pi}{17}\)

Step3: Find the second - smallest positive \(\theta\)

When \(k = 0\), \(\theta=\frac{\pi}{51}\)
When \(k = 1\), \(\theta=\frac{\pi}{51}+\frac{2\pi}{17}=\frac{\pi + 6\pi}{51}=\frac{7\pi}{51}\)

Answer:

\(r = 2^{\frac{1}{17}}\) and \(\theta=\frac{7\pi}{51}\)