Answer to Question #214413 in Complex Analysis for prince

Question

(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.

Answer 1

(4.1)

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

$$i^{145}=(i^4)^{36}i=i$$

(4.2)

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

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

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

$$=e^{i{\pi \over 2}}=\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}=i$$

$$\dfrac{z_1}{z_2}=\dfrac{\dfrac{\sqrt{2}}{2}e^{i{3\pi \over 4}}}{e^{i{\pi \over 2}}}=\dfrac{\sqrt{2}}{2}e^{i{\pi \over 4}}$$

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

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

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

(4.3)

$$z_1=-i=\cos(-\dfrac{\pi}{2})+i\sin(-\dfrac{\pi}{2})$$

$$z_2=-1-i\sqrt{3}=2(\cos(-\dfrac{2\pi}{3})+i\sin(-\dfrac{2\pi}{3}))$$

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

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

$$(z_1)^2=\cos(-\pi)+i\sin(-\pi))$$

$$(z_2)^{-1}=\dfrac{1}{2}(\cos(\dfrac{2\pi}{3})+i\sin(\dfrac{2\pi}{3}))$$

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

(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)

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

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

$z^4=w^3$

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

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

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

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

(4.4)

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

$z=i$

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

$$\lambda=-7i$$