Answer to Question #218688 in Linear Algebra for Tshego

Question

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.

Answer 1

(1)

$$i^{2666}=(i^4)^{666}i^2=-1$$ $$i^{145}=(i^4)^{36}i=i$$

(2)

$$z_1=\dfrac{-i}{-1+i}=e^{i{3\pi \over 2}}(\dfrac{\sqrt{2}}{2}e^{-i{3\pi \over 4}})\\ =\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}=\dfrac{\sqrt{2}}{2}(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4})=-\dfrac{1}{2}+i(\dfrac{1}{2})\\ z_2=\dfrac{1+i}{1-i}=\sqrt{2}e^{i{\pi \over 4}}(\dfrac{\sqrt{2}}{2}e^{i{\pi \over 4}})\\ =e^{i{\pi \over 2}}=\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}=i$$

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

Part i

$$\dfrac{z_1*z_3}{z_2}=\dfrac{\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}}{e^{i{\pi \over 2}}}=\dfrac{\sqrt{2}}{2}e^{i{\pi \over 4}}\\ =\dfrac{\sqrt{2}}{2}(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4})=\dfrac{1}{2}+i(\dfrac{1}{2})\\$$

Part ii

$\dfrac{z_1*z_2}{z_3}=\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}*{e^{i{\pi \over 2}}}=\dfrac{\sqrt{2}}{2}e^{i{5\pi \over 4}}\\ =\dfrac{\sqrt{2}}{2}(\cos\dfrac{5\pi}{4}+i\sin\dfrac{5\pi}{4})=-\dfrac{1}{2}-i(\dfrac{1}{2})\\$

Part iii

$\dfrac{z_2}{z_3*z_1}=\dfrac{e^{i{\pi \over 2}}}{\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}}=\sqrt{2}e^{-i{\pi \over 4}}\\ =\sqrt{2}(\cos(-\dfrac{\pi}{4})+i\sin(-\dfrac{\pi}{4})) =1-i$

(3)

$$z_1=-i=\cos(-\dfrac{\pi}{2})+i\sin(-\dfrac{\pi}{2})$$ $$z_2=-1-i\sqrt{3}=2(\cos(-\dfrac{2\pi}{3})+i\sin(-\dfrac{2\pi}{3}))$$ $$z_3=-\sqrt{3}+i=2(\cos(\dfrac{5\pi}{6})+i\sin(\dfrac{5\pi}{6}))$$ $$(z_3)^4=16(\cos(\dfrac{10\pi}{3})+i\sin(\dfrac{10\pi}{3}))$$ $$(z_1)^2=\cos(-\pi)+i\sin(-\pi))$$ $$(z_2)^{-1}=\dfrac{1}{2}(\cos(\dfrac{2\pi}{3})+i\sin(\dfrac{2\pi}{3}))$$ $$\dfrac{(z_3)^4}{(z_1)^2}\cdot(z_2)^{-1}=8(\cos(5\pi)+i\sin(5\pi))=-8$$

(a) $P(z)=z^2+a, a>0$

$$z^2+a=0=>z_1=-i\sqrt{a}, z_2=i\sqrt{a}$$

(b) $P(z)=z^3-z^2+z-1$

$$z^3-z^2+z-1=0$$ $$z^2(z-1)+(z-1)=0$$ $$z_1=1, z_2=-i, z_3=i$$

(c) $z^3-1=0$

$$(z-1)(z^2+z+1)=0$$ $$z_1=1, z_{2,3}=\dfrac{-1\pm i\sqrt{3}}{2}$$ $$z_1=1, z_2=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}, z_3=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$

(d)

$$8-8i=8\sqrt{2}(\cos(-\dfrac{\pi}{4})+i\sin(-\dfrac{\pi}{4}))$$ $$k=0: 2^{7/6}(\cos(-\dfrac{\pi}{12})+i\sin(-\dfrac{\pi}{12}))$$ $$k=1: 2^{7/6}(\cos(\dfrac{7\pi}{12})+i\sin(\dfrac{7\pi}{12}))$$ $$k=2: 2^{7/6}(\cos(\dfrac{5\pi}{4})+i\sin(\dfrac{5\pi}{4}))=-2^{2/3}-i(2^{2/3})$$

(e)

$$w=1+i=\sqrt{2}(\cos(\dfrac{\pi}{4})+i\sin(\dfrac{\pi}{4}))$$ $$w^3=2^{3/2}(\cos(\dfrac{3\pi}{4})+i\sin(\dfrac{3\pi}{4}))$$

$z^4=w^3$

$$k=0: 2^{3/8}(\cos(\dfrac{3\pi}{16})+i\sin(\dfrac{3\pi}{16}))$$ $$k=1: 2^{3/8}(\cos(\dfrac{11\pi}{16})+i\sin(\dfrac{11\pi}{16}))$$ $$k=2: 2^{3/8}(\cos(\dfrac{19\pi}{16})+i\sin(\dfrac{19\pi}{16}))$$ $$k=3: 2^{3/8}(\cos(\dfrac{27\pi}{16})+i\sin(\dfrac{27\pi}{16}))$$

(4)

$$P(z)=z^2+\lambda z-6$$

$z=i$

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