Answer to Question #218689 in Linear Algebra for Tshego

Question

Find the roots of the equation
1.)z^4 + 4 = 0 and z^4 - 4 = 0

2. ) Additional Exercises for practice are given below.
Find the roots of
(a) z^8-16 = 0
(b) z^8 + 16 = 0.

Answer 1

Solution.

1) $z^4 + 4 = 0$

$z^4 = -4$

$e^{i\pi} = -1$ and let $z = re^{i\theta}$

 $(re^{i\theta})^4 = 4e^{i\pi}$

$r^4 e^{i4\theta}= 4e^{i\pi}$

so$r^4 = 4 \implies r=\sqrt{2}$

$4\theta = \pi + 2n\pi \implies \theta = \frac{\pi}{4} + \frac{2n\pi}{4}$ where n=0,1,2,3

So, roots of the equation will be,

$z = \sqrt{2}e^{i(\frac{\pi}{4} )}, \sqrt{2}e^{i(\frac{\pi}{4} + \frac{\pi}{2})},\sqrt{2}e^{i(\frac{\pi}{4} + {\pi})},\sqrt{2}e^{i(\frac{\pi}{4} + \frac{3\pi}{2})}$

$z = \sqrt{2} e^{i(\frac{\pi}{4} )},\sqrt{2}e^{i(\frac{3\pi}{4} )},\sqrt{2}e^{i(\frac{5\pi}{4})},\sqrt{2}e^{i(\frac{7\pi}{4} )}$

Since, $e^{i\theta} = cos\theta + isin\theta$

$z_1 =\sqrt{2} e^{i(\frac{\pi}{4} )} = \sqrt{2}(cos\frac{\pi}{4} + isin\frac{\pi}{4}) = \sqrt{2}(\frac{1}{\sqrt{2} } +i\frac{1}{\sqrt{2}}) = 1+i$

$z_2 =\sqrt{2} e^{i(\frac{3\pi}{4} )} = \sqrt{2}(cos\frac{3\pi}{4} + isin\frac{3\pi}{4}) = \sqrt{2}(-\frac{1}{\sqrt{2} } +i\frac{1}{\sqrt{2}}) = -1+i$

$z_3 =\sqrt{2} e^{i(\frac{5\pi}{4} )} = \sqrt{2}(cos\frac{5\pi}{4} + isin\frac{5\pi}{4}) = \sqrt{2}(-\frac{1}{\sqrt{2} } -i\frac{1}{\sqrt{2}}) = -1-i$

$z_4 =\sqrt{2} e^{i(\frac{7\pi}{4} )} = \sqrt{2}(cos\frac{7\pi}{4} + isin\frac{7\pi}{4}) = \sqrt{2}(\frac{1}{\sqrt{2} } -i\frac{1}{\sqrt{2}}) = 1-i$

 $z^4-4 = 0 \implies (z^2-2)(z^2+2)=0$

$z^2-2=0 \implies z = \pm\sqrt{2}$

$z^2+2=0 \implies z^2 = - 2 \implies z=\pm\sqrt{2}i$

$z = \sqrt{2},-\sqrt{2},\sqrt{2}i,-\sqrt{2}i$

2) $z^8 - 16 = 0 \implies (z^4-4)(z^4+4) = 0$

$z = -\sqrt{2},\sqrt{2},-\sqrt{2}i,\sqrt{2}i,1+i,-1+i,-1-i,1-i$

 $z^8 + 16 = 0$

$z^8 = -16$

$(re^{i\theta})^8 = (16e^{i\pi})$

Solving it, $r = (16)^{1/8} \implies r=\sqrt{2}$

$8\theta = \pi + 2n\pi \implies \theta = \frac{\pi}{8} + \frac{2n\pi}{8 }$ where =0,1,2,3,4,5,6,7

$z_1 = \sqrt{2}e^{i\frac{\pi}{8}} = \sqrt{2}(cos\frac{\pi}{8}+isin\frac{\pi}{8})$

$z_2 = \sqrt{2}e^{i\frac{3\pi}{8}} = \sqrt{2}(cos\frac{3\pi}{8}+isin\frac{3\pi}{8})$

$z_3 = \sqrt{2}e^{i\frac{5\pi}{8}} = \sqrt{2}(cos\frac{5\pi}{8}+isin\frac{5\pi}{8})$

$z_4= \sqrt{2}e^{i\frac{7\pi}{8}} = \sqrt{2}(cos\frac{7\pi}{8}+isin\frac{7\pi}{8})$

$z_5 = \sqrt{2}e^{i\frac{9\pi}{8}} = \sqrt{2}(cos\frac{9\pi}{8}+isin\frac{9\pi}{8})$

$z_6 = \sqrt{2}e^{i\frac{11\pi}{8}} = \sqrt{2}(cos\frac{11\pi}{8}+isin\frac{11\pi}{8})$

$z_7 = \sqrt{2}e^{i\frac{13\pi}{8}} = \sqrt{2}(cos\frac{13\pi}{8}+isin\frac{13\pi}{8})$

$z_8 = \sqrt{2}e^{i\frac{15\pi}{8}} = \sqrt{2}(cos\frac{15\pi}{8}+isin\frac{15\pi}{8})$

Answer:

1)$z^4+4=0:$

$z_1 = 1+i$;

$z_2 =-1+i$;

$z_3 = -1-i$;

$z_4 =1-i$;

$z^4-4=0:$

$z = \sqrt{2},-\sqrt{2},\sqrt{2}i,-\sqrt{2}iz= 2 ​ ,− 2 ​ , 2 ​ i,− 2 ​ i$ ;

2)$z^8-16=0:$

$z=− 2 ​ , 2 ​ ,− 2 ​ i, 2 ​ i,1+i,−1+i,−1−i,1−i$ ;

$z^8+16=0:$

$z_1 = \sqrt{2}(cos\frac{\pi}{8}+isin\frac{\pi}{8});$

$z_2 = \sqrt{2}(cos\frac{3\pi}{8}+isin\frac{3\pi}{8});$

$z_3 = \sqrt{2}(cos\frac{5\pi}{8}+isin\frac{5\pi}{8}) ;$

$z_4= \sqrt{2}(cos\frac{7\pi}{8}+isin\frac{7\pi}{8}) ;$

$z_5 = \sqrt{2}(cos\frac{9\pi}{8}+isin\frac{9\pi}{8}) ;$

$z_6 = \sqrt{2}(cos\frac{11\pi}{8}+isin\frac{11\pi}{8}) ;$

$z_7 = \sqrt{2}(cos\frac{13\pi}{8}+isin\frac{13\pi}{8}) ;$

$z_8 = \sqrt{2}(cos\frac{15\pi}{8}+isin\frac{15\pi}{8}).$