Question

Find all z,such that z^(2015)=321

Answer 1

Answer on Question #55967 – Math – Complex Analysis

Question

Find all $z$, such that $z^{2015} = 321$.

Solution

Let $w = 312$, then $\rho = 321, \varphi = 0$. Then using the formula

$$\sqrt[n]{w} = \left\{ \sqrt[n]{\rho} e^{\frac{\varphi + 2\pi k}{n}}, k = 0, 1, \dots, n - 1 \right\} \text{ we obtain:}$$ $$z = \sqrt[2015]{321 + i \cdot 0} = \left\{ \sqrt[2015]{321} e^{\frac{2\pi k}{2015}}, k = 0, 1, \dots, 2014 \right\}, \text{ where } \sqrt[2015]{321} \approx 1.003.$$

Answer: $\left\{ \sqrt[2015]{321} e^{\frac{2\pi k}{2015}}, k = 0, 1, \dots, 2014 \right\}$.

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