Answer to Question #212667 in Algebra for Faith

Question

1.find the roots of the equation: z^4+4=0 and z^4-4=0.

2.additional exercises for practice are given below.find the roots of:

2.1.z^8-16=0.

2.2.z^8+16=0.

Answer 1

$z^4+4=0\\ z^4=-4\\ z^4=4(\cos(\pi)+i\sin(\pi))\\ z^4=4(\cos(2\pi{n}+\pi)+i\sin(2\pi{n}+\pi))\\ z=4^\frac{1}{4}(\cos(\frac{2\pi{n}+\pi}{4})+\\i\sin(\frac{2\pi{n}+\pi)}{4})\\ n=0,1,2,3\\ z_o=4^\frac{1}{4}(\cos(\frac{0+\pi}{4})+\\i\sin(\frac{0+\pi)}{4})\\ z_o=4^\frac{1}{4}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))\\ z_1=4^\frac{1}{4}(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))\\ z_2=4^\frac{1}{4}(\cos(\frac{5\pi}{4})+i\sin(\frac{5\pi}{4}))\\ z_3=4^\frac{1}{4}(\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4}))\\$

To solve $z^4-4=0$

$z^4=4\\ z^4 =4(cos(0)+isin(0))\\ z^4=4(\cos(2\pi{n})+i\sin(2\pi{n})\\ z=4^\frac{1}{4}(\cos(\frac{2\pi{n}}{4})+\\i\sin(\frac{2\pi{n})}{4})\\ z_o=4^\frac{1}{4}(\cos(0)+i\sin(0))\\ z_o=4^\frac{1}{4}\\ z_1=4^\frac{1}{4}(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))\\ z_2=4^\frac{1}{4}(\cos(\pi)+i\sin(\pi))\\ z_3=4^\frac{1}{4}(\cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2}))\\$

To solve $z^4-16=0$

$z^4=16\\ z^4 =16(cos(0)+isin(0))\\ z^4=16(\cos(2\pi{n})+i\sin(2\pi{n})\\ z=16^\frac{1}{4}(\cos(\frac{2\pi{n}}{4})+\\i\sin(\frac{2\pi{n})}{4})\\ z_o=16^\frac{1}{4}(\cos(0)+i\sin(0))\\ z_o=16^\frac{1}{4}=2\\ z_1=2(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))\\ z_2=2(\cos(\pi)+i\sin(\pi))\\ z_3=2(\cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2}))\\$

To solve $z^4+16=0$

$z^4=-16\\ z^4=16(\cos(2\pi{n}+\pi)+i\sin(2\pi{n}+\pi))\\ z=16^\frac{1}{4}(\cos(\frac{2\pi{n}+\pi}{4})+\\i\sin(\frac{2\pi{n}+\pi)}{4})\\ n=0,1,2,3\\ z_o=16^\frac{1}{4}(\cos(\frac{0+\pi}{4})+\\i\sin(\frac{0+\pi)}{4})\\ z_o=2(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))=-2\\ z_1=2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))\\ z_2=2(\cos(\frac{5\pi}{4})+i\sin(\frac{5\pi}{4}))\\ z_3=2(\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4}))\\$