Question #214413

(4.1) Determine the complex numbers i2666 and i145.

(4.2) Let z1 = (6) −i −1+i and z2 = 1+i 1−i . Express z1z3/z2 , z1z2/z3 , and z1/z3z2 in both polar and standard forms.

(4.3) Additional Exercises for practice: Express z1 = −i, z2 = −1 − i √ 3, and z3= − √ 3 + i in polar form and use your results to find z43 /z21 z-12. Find the roots of the polynomials below.

(a) P(z) = z2 + a for a > 0

(b) P(z) = z3 − z2 + z − 1.

(c) Find the roots of z (4) 3 − 1

(d) Find in standard forms, the cube roots of 8 − 8i

(e) Let w = 1 + i. Solve for the complex number z from the equation z4 = w3 . (4.4) Find the value(s) for λ so that α = i is a root of P(z) = z2 + λz − 6.


1
Expert's answer
2021-07-08T09:55:37-0400

(4.1)


i2666=(i4)666i2=1i^{2666}=(i^4)^{666}i^2=-1

i145=(i4)36i=ii^{145}=(i^4)^{36}i=i

(4.2)


z1=i1+i=ei3π2(22ei3π4)z_1=\dfrac{-i}{-1+i}=e^{i{3\pi \over 2}}(\dfrac{\sqrt{2}}{2}e^{-i{3\pi \over 4}})

=22ei3π4=22(cos3π4+isin3π4)=12+i(12)=\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}=\dfrac{\sqrt{2}}{2}(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4})=-\dfrac{1}{2}+i(\dfrac{1}{2})


z2=1+i1i=2eiπ4(22eiπ4)z_2=\dfrac{1+i}{1-i}=\sqrt{2}e^{i{\pi \over 4}}(\dfrac{\sqrt{2}}{2}e^{i{\pi \over 4}})

=eiπ2=cosπ2+isinπ2=i=e^{i{\pi \over 2}}=\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}=i

z1z2=22ei3π4eiπ2=22eiπ4\dfrac{z_1}{z_2}=\dfrac{\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}}{e^{i{\pi \over 2}}}=\dfrac{\sqrt{2}}{2}e^{i{\pi \over 4}}

=22(cosπ4+isinπ4)=12+i(12)=\dfrac{\sqrt{2}}{2}(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4})=\dfrac{1}{2}+i(\dfrac{1}{2})

z2z1=eiπ222ei3π4=2eiπ4\dfrac{z_2}{z_1}=\dfrac{e^{i{\pi \over 2}}}{\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}}=\sqrt{2}e^{-i{\pi \over 4}}

=2(cos(π4)+isin(π4))=1i=\sqrt{2}(\cos(-\dfrac{\pi}{4})+i\sin(-\dfrac{\pi}{4}))=1-i

(4.3)


z1=i=cos(π2)+isin(π2)z_1=-i=\cos(-\dfrac{\pi}{2})+i\sin(-\dfrac{\pi}{2})

z2=1i3=2(cos(2π3)+isin(2π3))z_2=-1-i\sqrt{3}=2(\cos(-\dfrac{2\pi}{3})+i\sin(-\dfrac{2\pi}{3}))

z3=3+i=2(cos(5π6)+isin(5π6))z_3=-\sqrt{3}+i=2(\cos(\dfrac{5\pi}{6})+i\sin(\dfrac{5\pi}{6}))

(z3)4=16(cos(10π3)+isin(10π3))(z_3)^4=16(\cos(\dfrac{10\pi}{3})+i\sin(\dfrac{10\pi}{3}))

(z1)2=cos(π)+isin(π))(z_1)^2=\cos(-\pi)+i\sin(-\pi))

(z2)1=12(cos(2π3)+isin(2π3))(z_2)^{-1}=\dfrac{1}{2}(\cos(\dfrac{2\pi}{3})+i\sin(\dfrac{2\pi}{3}))

(z3)4(z1)2(z2)1=8(cos(5π)+isin(5π))=8\dfrac{(z_3)^4}{(z_1)^2}\cdot(z_2)^{-1}=8(\cos(5\pi)+i\sin(5\pi))=-8

(a) P(z)=z2+a,a>0P(z)=z^2+a, a>0


z2+a=0=>z1=ia,z2=iaz^2+a=0=>z_1=-i\sqrt{a}, z_2=i\sqrt{a}

(b) P(z)=z3z2+z1P(z)=z^3-z^2+z-1


z3z2+z1=0z^3-z^2+z-1=0

z2(z1)+(z1)=0z^2(z-1)+(z-1)=0

z1=1,z2=i,z3=iz_1=1, z_2=-i, z_3=i

(c) z31=0z^3-1=0


(z1)(z2+z+1)=0(z-1)(z^2+z+1)=0

z1=1,z2,3=1±i32z_1=1, z_{2,3}=\dfrac{-1\pm i\sqrt{3}}{2}

z1=1,z2=12i32,z3=12+i32z_1=1, z_2=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}, z_3=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}

(d)


88i=82(cos(π4)+isin(π4))8-8i=8\sqrt{2}(\cos(-\dfrac{\pi}{4})+i\sin(-\dfrac{\pi}{4}))

k=0:27/6(cos(π12)+isin(π12))k=0: 2^{7/6}(\cos(-\dfrac{\pi}{12})+i\sin(-\dfrac{\pi}{12}))

k=1:27/6(cos(7π12)+isin(7π12))k=1: 2^{7/6}(\cos(\dfrac{7\pi}{12})+i\sin(\dfrac{7\pi}{12}))

k=2:27/6(cos(5π4)+isin(5π4))=22/3i(22/3)k=2: 2^{7/6}(\cos(\dfrac{5\pi}{4})+i\sin(\dfrac{5\pi}{4}))=-2^{2/3}-i(2^{2/3})

(e)


w=1+i=2(cos(π4)+isin(π4))w=1+i=\sqrt{2}(\cos(\dfrac{\pi}{4})+i\sin(\dfrac{\pi}{4}))

w3=23/2(cos(3π4)+isin(3π4))w^3=2^{3/2}(\cos(\dfrac{3\pi}{4})+i\sin(\dfrac{3\pi}{4}))

z4=w3z^4=w^3


k=0:23/8(cos(3π16)+isin(3π16))k=0: 2^{3/8}(\cos(\dfrac{3\pi}{16})+i\sin(\dfrac{3\pi}{16}))

k=1:23/8(cos(11π16)+isin(11π16))k=1: 2^{3/8}(\cos(\dfrac{11\pi}{16})+i\sin(\dfrac{11\pi}{16}))

k=2:23/8(cos(19π16)+isin(19π16))k=2: 2^{3/8}(\cos(\dfrac{19\pi}{16})+i\sin(\dfrac{19\pi}{16}))

k=3:23/8(cos(27π16)+isin(27π16))k=3: 2^{3/8}(\cos(\dfrac{27\pi}{16})+i\sin(\dfrac{27\pi}{16}))

(4.4)


P(z)=z2+λz6P(z)=z^2+\lambda z-6

z=iz=i


(i)2+λi6=0(i)^2+\lambda i-6=0

λ=7i\lambda=-7i




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