Answer to Question #218689 in Linear Algebra for Tshego

Question #218689
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.
1
Expert's answer
2021-07-20T17:24:47-0400

Solution.

1) z4+4=0z^4 + 4 = 0

z4=4z^4 = -4

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

 (reiθ)4=4eiπ(re^{i\theta})^4 = 4e^{i\pi}

r4ei4θ=4eiπr^4 e^{i4\theta}= 4e^{i\pi}

sor4=4    r=2r^4 = 4 \implies r=\sqrt{2}

4θ=π+2nπ    θ=π4+2nπ44\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=2ei(π4),2ei(π4+π2),2ei(π4+π),2ei(π4+3π2)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=2ei(π4),2ei(3π4),2ei(5π4),2ei(7π4)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, eiθ=cosθ+isinθe^{i\theta} = cos\theta + isin\theta

z1=2ei(π4)=2(cosπ4+isinπ4)=2(12+i12)=1+iz_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

z2=2ei(3π4)=2(cos3π4+isin3π4)=2(12+i12)=1+iz_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

z3=2ei(5π4)=2(cos5π4+isin5π4)=2(12i12)=1iz_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

z4=2ei(7π4)=2(cos7π4+isin7π4)=2(12i12)=1iz_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

 z44=0    (z22)(z2+2)=0z^4-4 = 0 \implies (z^2-2)(z^2+2)=0

z22=0    z=±2z^2-2=0 \implies z = \pm\sqrt{2}

z2+2=0    z2=2    z=±2iz^2+2=0 \implies z^2 = - 2 \implies z=\pm\sqrt{2}i

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

2) z816=0    (z44)(z4+4)=0z^8 - 16 = 0 \implies (z^4-4)(z^4+4) = 0

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

 z8+16=0z^8 + 16 = 0

z8=16z^8 = -16

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

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

8θ=π+2nπ    θ=π8+2nπ88\theta = \pi + 2n\pi \implies \theta = \frac{\pi}{8} + \frac{2n\pi}{8 } where =0,1,2,3,4,5,6,7

z1=2eiπ8=2(cosπ8+isinπ8)z_1 = \sqrt{2}e^{i\frac{\pi}{8}} = \sqrt{2}(cos\frac{\pi}{8}+isin\frac{\pi}{8})

z2=2ei3π8=2(cos3π8+isin3π8)z_2 = \sqrt{2}e^{i\frac{3\pi}{8}} = \sqrt{2}(cos\frac{3\pi}{8}+isin\frac{3\pi}{8})

z3=2ei5π8=2(cos5π8+isin5π8)z_3 = \sqrt{2}e^{i\frac{5\pi}{8}} = \sqrt{2}(cos\frac{5\pi}{8}+isin\frac{5\pi}{8})

z4=2ei7π8=2(cos7π8+isin7π8)z_4= \sqrt{2}e^{i\frac{7\pi}{8}} = \sqrt{2}(cos\frac{7\pi}{8}+isin\frac{7\pi}{8})

z5=2ei9π8=2(cos9π8+isin9π8)z_5 = \sqrt{2}e^{i\frac{9\pi}{8}} = \sqrt{2}(cos\frac{9\pi}{8}+isin\frac{9\pi}{8})

z6=2ei11π8=2(cos11π8+isin11π8)z_6 = \sqrt{2}e^{i\frac{11\pi}{8}} = \sqrt{2}(cos\frac{11\pi}{8}+isin\frac{11\pi}{8})

z7=2ei13π8=2(cos13π8+isin13π8)z_7 = \sqrt{2}e^{i\frac{13\pi}{8}} = \sqrt{2}(cos\frac{13\pi}{8}+isin\frac{13\pi}{8})

z8=2ei15π8=2(cos15π8+isin15π8)z_8 = \sqrt{2}e^{i\frac{15\pi}{8}} = \sqrt{2}(cos\frac{15\pi}{8}+isin\frac{15\pi}{8})

Answer:

1)z4+4=0:z^4+4=0:

z1=1+iz_1 = 1+i;

z2=1+iz_2 =-1+i;

z3=1iz_3 = -1-i;

z4=1iz_4 =1-i;

z44=0:z^4-4=0:

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

2)z816=0:z^8-16=0:

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

z8+16=0:z^8+16=0:

z1=2(cosπ8+isinπ8);z_1 = \sqrt{2}(cos\frac{\pi}{8}+isin\frac{\pi}{8});

z2=2(cos3π8+isin3π8);z_2 = \sqrt{2}(cos\frac{3\pi}{8}+isin\frac{3\pi}{8});

z3=2(cos5π8+isin5π8);z_3 = \sqrt{2}(cos\frac{5\pi}{8}+isin\frac{5\pi}{8}) ;

z4=2(cos7π8+isin7π8);z_4= \sqrt{2}(cos\frac{7\pi}{8}+isin\frac{7\pi}{8}) ;

z5=2(cos9π8+isin9π8);z_5 = \sqrt{2}(cos\frac{9\pi}{8}+isin\frac{9\pi}{8}) ;

z6=2(cos11π8+isin11π8);z_6 = \sqrt{2}(cos\frac{11\pi}{8}+isin\frac{11\pi}{8}) ;

z7=2(cos13π8+isin13π8);z_7 = \sqrt{2}(cos\frac{13\pi}{8}+isin\frac{13\pi}{8}) ;

z8=2(cos15π8+isin15π8).z_8 = \sqrt{2}(cos\frac{15\pi}{8}+isin\frac{15\pi}{8}).

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