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Roots of Complex Numbers Calculator

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Example
Created on 2024-06-20Asked by Luna Martinez (Solvelet student)
Find all the cube roots of the complex number z=8+8iz = 8 + 8i.

Solution

To find all the cube roots of the complex number z=8+8iz = 8 + 8i: 1. **Express zz in polar form:** z=8+8i=82(cosπ4+isinπ4). z = 8 + 8i = 8\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right). 2. **Apply De Moivre's Theorem for cube roots:** z1/3=(82)1/3(cosπ/4+2kπ3+isinπ/4+2kπ3),k=0,1,2. z^{1/3} = \left(8\sqrt{2}\right)^{1/3} \left(\cos\frac{\pi/4 + 2k\pi}{3} + i\sin\frac{\pi/4 + 2k\pi}{3}\right), \quad k = 0, 1, 2. 3. **Calculate the cube roots:** (82)1/3=223. \left(8\sqrt{2}\right)^{1/3} = 2\sqrt[3]{2}. For k=0k = 0: cosπ/12+isinπ/12. \cos\frac{\pi/12} + i\sin\frac{\pi/12}. For k=1k = 1: cos9π/12+isin9π/12. \cos\frac{9\pi/12} + i\sin\frac{9\pi/12}. For k=2k = 2: cos17π/12+isin17π/12. \cos\frac{17\pi/12} + i\sin\frac{17\pi/12}. 4. **Result:** The three cube roots are: 223(cosπ12+isinπ12), 2\sqrt[3]{2}\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right), 223(cos3π4+isin3π4), 2\sqrt[3]{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right), 223(cos17π12+isin17π12). 2\sqrt[3]{2}\left(\cos\frac{17\pi}{12} + i\sin\frac{17\pi}{12}\right). Solved on Solvelet with Basic AI Model
Some of the related questions asked by Alexander Sanchez on Solvelet
1. Find all the roots of the equation z3=8 z^3 = 8 in the complex plane.2. Determine the roots of the equation z4=16 z^4 = -16 in the complex plane.,
DefinitionThe roots of a complex number are the solutions to equations of the form zn=w where z, w are complex numbers. The roots can be obtained using De Moivre's method as follows (z_k = Solution Roots of Complex numbers Finding using De moivre's Formula.
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