Question #218688
1.) Determine the complex numbers i^2666 and i^145

2.) Let z1 =-i/-1+I and z2 =1+i/ 1- i. Express z1z3/z2, z1z2/z3, and z1/z3z2 in both polar and standard forms.

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
(z4/3)
/z2/1 z -1/ 2 .
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 z3-1
(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 z^4 = w3.

4.)Find the value(s) for λ so that α = i is a root of P(z) = z^2 + λz-6.
1
Expert's answer
2021-07-21T07:01:16-0400

(1)

i2666=(i4)666i2=1i^{2666}=(i^4)^{666}i^2=-1i145=(i4)36i=ii^{145}=(i^4)^{36}i=i

(2)

z1=i1+i=ei3π2(22ei3π4)=22ei3π4=22(cos3π4+isin3π4)=12+i(12)z2=1+i1i=2eiπ4(22eiπ4)=eiπ2=cosπ2+isinπ2=iz_1=\dfrac{-i}{-1+i}=e^{i{3\pi \over 2}}(\dfrac{\sqrt{2}}{2}e^{-i{3\pi \over 4}})\\ =\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})\\ z_2=\dfrac{1+i}{1-i}=\sqrt{2}e^{i{\pi \over 4}}(\dfrac{\sqrt{2}}{2}e^{i{\pi \over 4}})\\ =e^{i{\pi \over 2}}=\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}=i


Since Z3 is not given, we will assume that it is equal to 1

Part i

z1z3z2=22ei3π4eiπ2=22eiπ4=22(cosπ4+isinπ4)=12+i(12)\dfrac{z_1*z_3}{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}}\\ =\dfrac{\sqrt{2}}{2}(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4})=\dfrac{1}{2}+i(\dfrac{1}{2})\\


Part ii

z1z2z3=22ei3π4eiπ2=22ei5π4=22(cos5π4+isin5π4)=12i(12)\dfrac{z_1*z_2}{z_3}=\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}*{e^{i{\pi \over 2}}}=\dfrac{\sqrt{2}}{2}e^{i{5\pi \over 4}}\\ =\dfrac{\sqrt{2}}{2}(\cos\dfrac{5\pi}{4}+i\sin\dfrac{5\pi}{4})=-\dfrac{1}{2}-i(\dfrac{1}{2})\\


Part iii

z2z3z1=eiπ222ei3π4=2eiπ4=2(cos(π4)+isin(π4))=1i\dfrac{z_2}{z_3*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}}\\ =\sqrt{2}(\cos(-\dfrac{\pi}{4})+i\sin(-\dfrac{\pi}{4})) =1-i

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



P(z)=z2+λz6P(z)=z^2+\lambda z-6

z=iz=i



(i)2+λi6=0(i)^2+\lambda i-6=0λ=7i\lambda=-7i

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