9.1 The Moivre’s Theorem
z = r ( cos θ + i sin θ ) z n = r n ( cos n θ + i sin n θ ) , n ∈ N z=r(\cos\theta+i\sin \theta)\\
z^n=r^n(\cos n\theta+i\sin n\theta), n\in N z = r ( cos θ + i sin θ ) z n = r n ( cos n θ + i sin n θ ) , n ∈ N
9.2 Let r = 1 r=1 r = 1
z = cos θ + i sin θ z 5 = cos 5 θ + i sin 5 θ = ( cos θ + i sin θ ) 5 = = cos 5 θ + 5 i cos 4 θ sin θ + 10 i 2 cos 3 θ sin 2 θ + + 10 i 3 cos 2 θ sin 3 θ + 5 i 4 cos θ sin 4 θ + i 5 sin 5 θ = = cos 5 θ − 10 cos 3 θ sin 2 θ + 5 cos θ sin 4 θ + + i ( 5 cos 4 θ sin θ − 10 cos 2 θ sin 3 θ + sin 5 θ ) cos 5 θ = cos 5 θ − 10 cos 3 θ sin 2 θ + 5 cos θ sin 4 θ z=\cos\theta+i\sin\theta\\
z^5=\cos5\theta+i\sin5\theta=(\cos\theta+i\sin\theta)^5=\\
=\cos^5\theta+5i\cos^4\theta\sin\theta+10i^2\cos^3\theta\sin^2\theta+\\
+10i^3\cos^2\theta\sin^3\theta+5i^4\cos\theta\sin^4\theta+i^5\sin^5\theta=\\
=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta+\\
+i(5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta)\\
\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\\ z = cos θ + i sin θ z 5 = cos 5 θ + i sin 5 θ = ( cos θ + i sin θ ) 5 = = cos 5 θ + 5 i cos 4 θ sin θ + 10 i 2 cos 3 θ sin 2 θ + + 10 i 3 cos 2 θ sin 3 θ + 5 i 4 cos θ sin 4 θ + i 5 sin 5 θ = = cos 5 θ − 10 cos 3 θ sin 2 θ + 5 cos θ sin 4 θ + + i ( 5 cos 4 θ sin θ − 10 cos 2 θ sin 3 θ + sin 5 θ ) cos 5 θ = cos 5 θ − 10 cos 3 θ sin 2 θ + 5 cos θ sin 4 θ
z 4 = cos 4 θ + i sin 4 θ = ( cos θ + i sin θ ) 4 = = cos 4 θ + 4 i cos 3 θ sin θ + 6 i 2 cos 2 θ sin 2 θ + + 6 i 3 cos θ sin 3 θ + i 4 sin 4 θ = = cos 4 θ − 6 cos 2 θ sin 2 θ + sin 4 θ + + i ( 4 cos 3 θ sin θ − 6 cos θ sin 3 θ ) sin 4 θ = 4 cos 3 θ sin θ − 6 cos θ sin 3 θ z^4=\cos4\theta+i\sin4\theta=(\cos\theta+i\sin\theta)^4=\\
=\cos^4\theta+4i\cos^3\theta\sin\theta+6i^2\cos^2\theta\sin^2\theta+\\
+6i^3\cos\theta\sin^3\theta+i^4\sin^4\theta=\\
=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta+\\
+i(4\cos^3\theta\sin\theta-6\cos\theta\sin^3\theta)\\
\sin4\theta=4\cos^3\theta\sin\theta-6\cos\theta\sin^3\theta z 4 = cos 4 θ + i sin 4 θ = ( cos θ + i sin θ ) 4 = = cos 4 θ + 4 i cos 3 θ sin θ + 6 i 2 cos 2 θ sin 2 θ + + 6 i 3 cos θ sin 3 θ + i 4 sin 4 θ = = cos 4 θ − 6 cos 2 θ sin 2 θ + sin 4 θ + + i ( 4 cos 3 θ sin θ − 6 cos θ sin 3 θ ) sin 4 θ = 4 cos 3 θ sin θ − 6 cos θ sin 3 θ
i 2 = − 1 , i 3 = − i , i 4 = 1 , i 5 = i i^2=-1, i^3=-i, i^4=1,i^5=i i 2 = − 1 , i 3 = − i , i 4 = 1 , i 5 = i
9.3 a)
z = w 6 , w < 0 w = ρ ( cos θ + i sin θ ) z = ρ 6 ( cos θ + 2 π k 6 + i sin θ + 2 π k 6 ) , k = 0 , 1 , 2 , 3 , 4 , 5 z=\sqrt[6]{w}, w<0\\
w=\rho(\cos\theta+i\sin\theta)\\
z=\sqrt[6]\rho(\cos\frac{\theta+2\pi k}{6}+i\sin\frac{\theta+2\pi k}{6}), k=0,1,2,3,4,5 z = 6 w , w < 0 w = ρ ( cos θ + i sin θ ) z = 6 ρ ( cos 6 θ + 2 πk + i sin 6 θ + 2 πk ) , k = 0 , 1 , 2 , 3 , 4 , 5
b)
− 729 6 = − 1 ⋅ 3 6 6 = 3 − 1 6 − 1 = cos π + i sin π − 1 6 = cos π + 2 π k 6 + sin π + 2 π k 6 , k = 0 , 1 , 2 , 3 , 4 , 5 − 729 6 = 3 ( cos π + 2 π k 6 + sin π + 2 π k 6 ) , k = 0 , 1 , 2 , 3 , 4 , 5 \sqrt[6]{-729}=\sqrt[6]{-1\cdot3^6}=3\sqrt[6]{-1}\\
-1=\cos\pi+i\sin\pi\\
\sqrt[6]{-1}=\cos\frac{\pi+2\pi k}{6}+\sin\frac{\pi+2\pi k}{6}, k=0,1,2,3,4,5\\
\sqrt[6]{-729}=3(\cos\frac{\pi+2\pi k}{6}+\sin\frac{\pi+2\pi k}{6}),\\
k=0,1,2,3,4,5\\ 6 − 729 = 6 − 1 ⋅ 3 6 = 3 6 − 1 − 1 = cos π + i sin π 6 − 1 = cos 6 π + 2 πk + sin 6 π + 2 πk , k = 0 , 1 , 2 , 3 , 4 , 5 6 − 729 = 3 ( cos 6 π + 2 πk + sin 6 π + 2 πk ) , k = 0 , 1 , 2 , 3 , 4 , 5
− 729 6 = 3 ( cos π 6 + sin π 6 ) = 3 ( 3 2 + i 1 2 ) , − 729 6 = 3 ( cos 3 π 6 + sin 3 π 6 ) = 3 i , − 729 6 = 3 ( cos 5 π 6 + sin 5 π 6 ) = 3 ( − 3 2 + i 1 2 ) , − 729 6 = 3 ( cos 7 π 6 + sin 7 π 6 ) = 3 ( − 3 2 − i 1 2 ) , − 729 6 = 3 ( cos 9 π 6 + sin 9 π 6 ) = − 3 i , − 729 6 = 3 ( cos 11 π 6 + sin 11 π 6 ) = 3 ( 3 2 − i 1 2 ) \sqrt[6]{-729}=3(\cos\frac{\pi}{6}+\sin\frac{\pi}{6})=3(\frac{\sqrt3}{2}+i\frac{1}{2}),\\
\sqrt[6]{-729}=3(\cos\frac{3\pi }{6}+\sin\frac{3\pi }{6})=3i,\\
\sqrt[6]{-729}=3(\cos\frac{5\pi }{6}+\sin\frac{5\pi }{6})=3(-\frac{\sqrt3}{2}+i\frac{1}{2}),\\
\sqrt[6]{-729}=3(\cos\frac{7\pi}{6}+\sin\frac{7\pi}{6})=3(-\frac{\sqrt3}{2}-i\frac{1}{2}),\\
\sqrt[6]{-729}=3(\cos\frac{9\pi}{6}+\sin\frac{9\pi}{6})=-3i,\\
\sqrt[6]{-729}=3(\cos\frac{11\pi }{6}+\sin\frac{11\pi }{6})=3(\frac{\sqrt3}{2}-i\frac{1}{2})\\ 6 − 729 = 3 ( cos 6 π + sin 6 π ) = 3 ( 2 3 + i 2 1 ) , 6 − 729 = 3 ( cos 6 3 π + sin 6 3 π ) = 3 i , 6 − 729 = 3 ( cos 6 5 π + sin 6 5 π ) = 3 ( − 2 3 + i 2 1 ) , 6 − 729 = 3 ( cos 6 7 π + sin 6 7 π ) = 3 ( − 2 3 − i 2 1 ) , 6 − 729 = 3 ( cos 6 9 π + sin 6 9 π ) = − 3 i , 6 − 729 = 3 ( cos 6 11 π + sin 6 11 π ) = 3 ( 2 3 − i 2 1 )
c)
z = cos θ + i sin θ u + i v = ( 1 + z ) ( 1 + z 2 ) = = ( ( 1 + cos θ ) + i sin θ ) ⋅ ( 1 + cos 2 θ + 2 i cos θ sin θ − sin 2 θ ) = = ( ( 1 + cos θ ) + i sin θ ) ⋅ ( ( 1 + cos 2 θ − sin 2 θ ) + 2 i cos θ sin θ ) = = ( ( 1 + cos θ ) + i sin θ ) ⋅ ( 2 cos 2 θ + 2 i cos θ sin θ ) = = 2 cos 2 θ ( 1 + cos θ ) − 2 cos θ sin 2 θ + + i ( 2 cos θ sin θ ( 1 + cos θ ) + 2 sin θ cos 2 θ ) u = 2 cos 2 θ ( 1 + cos θ ) − 2 cos θ sin 2 θ = = 2 cos θ ( cos θ + cos 2 θ ) v = 2 cos θ sin θ ( 1 + cos θ ) + 2 sin θ cos 2 θ = = 2 sin θ cos θ ( 1 + 2 cos θ ) = 2 cos θ ( sin θ + sin 2 θ ) u tan 3 θ 2 = 2 cos θ ( cos θ + cos 2 θ ) sin 3 θ 2 cos 3 θ 2 = = 2 cos θ cos 3 θ 2 sin θ 2 + sin 5 θ 2 − sin θ 2 + sin 7 θ 2 2 = = 2 cos θ cos 3 θ 2 2 sin 3 θ cos θ 2 2 = = 2 cos θ ( 2 sin 3 θ 2 cos θ 2 ) = = 2 cos θ ( sin θ + sin 2 θ ) = v z=\cos\theta+i\sin\theta\\
u+iv=(1+z)(1+z^2)=\\
=((1+\cos\theta)+i\sin\theta)\cdot\\(1+\cos^2\theta+2i\cos\theta\sin\theta-\sin^2\theta)=\\
=((1+\cos\theta)+i\sin\theta)\cdot\\((1+\cos^2\theta-\sin^2\theta)+2i\cos\theta\sin\theta)=\\
=((1+\cos\theta)+i\sin\theta)\cdot\\(2\cos^2\theta+2i\cos\theta\sin\theta)=\\
=2\cos^2\theta(1+\cos\theta)-2\cos\theta\sin^2\theta+\\
+i(2\cos\theta\sin\theta(1+\cos\theta)+2\sin\theta\cos^2\theta)\\
u=2\cos^2\theta(1+\cos\theta)-2\cos\theta\sin^2\theta=\\
=2\cos\theta(\cos\theta+\cos2\theta)\\
v=2\cos\theta\sin\theta(1+\cos\theta)+2\sin\theta\cos^2\theta=\\
=2\sin\theta\cos\theta(1+2\cos\theta)=2\cos\theta(\sin\theta+\sin2\theta)\\
u\tan\frac{3\theta}{2}=2\cos\theta(\cos\theta+\cos2\theta)\frac{\sin\frac{3\theta}{2}}{\cos\frac{3\theta}{2}}=\\
=\frac{2\cos\theta}{\cos\frac{3\theta}{2}}\frac{\sin\frac{\theta}{2}+\sin\frac{5\theta}{2}-\sin\frac{\theta}{2}+\sin\frac{7\theta}{2}}{2}=\\
=\frac{2\cos\theta}{\cos\frac{3\theta}{2}}\frac{2\sin3\theta\cos\frac{\theta}{2}}{2}=\\
=2\cos\theta(2\sin\frac{3\theta}{2}\cos\frac{\theta}{2})=\\
=2\cos\theta(\sin\theta+\sin2\theta)=v z = cos θ + i sin θ u + i v = ( 1 + z ) ( 1 + z 2 ) = = (( 1 + cos θ ) + i sin θ ) ⋅ ( 1 + cos 2 θ + 2 i cos θ sin θ − sin 2 θ ) = = (( 1 + cos θ ) + i sin θ ) ⋅ (( 1 + cos 2 θ − sin 2 θ ) + 2 i cos θ sin θ ) = = (( 1 + cos θ ) + i sin θ ) ⋅ ( 2 cos 2 θ + 2 i cos θ sin θ ) = = 2 cos 2 θ ( 1 + cos θ ) − 2 cos θ sin 2 θ + + i ( 2 cos θ sin θ ( 1 + cos θ ) + 2 sin θ cos 2 θ ) u = 2 cos 2 θ ( 1 + cos θ ) − 2 cos θ sin 2 θ = = 2 cos θ ( cos θ + cos 2 θ ) v = 2 cos θ sin θ ( 1 + cos θ ) + 2 sin θ cos 2 θ = = 2 sin θ cos θ ( 1 + 2 cos θ ) = 2 cos θ ( sin θ + sin 2 θ ) u tan 2 3 θ = 2 cos θ ( cos θ + cos 2 θ ) c o s 2 3 θ s i n 2 3 θ = = c o s 2 3 θ 2 c o s θ 2 s i n 2 θ + s i n 2 5 θ − s i n 2 θ + s i n 2 7 θ = = c o s 2 3 θ 2 c o s θ 2 2 s i n 3 θ c o s 2 θ = = 2 cos θ ( 2 sin 2 3 θ cos 2 θ ) = = 2 cos θ ( sin θ + sin 2 θ ) = v
u 2 + v 2 = u = 4 cos 2 θ ( cos θ + cos 2 θ ) 2 + + 4 cos 2 θ ( sin θ + sin 2 θ ) 2 = = 4 cos 2 θ ( cos 2 θ + 2 cos θ cos 2 θ + cos 2 2 θ + + sin 2 θ + 2 sin θ sin 2 θ + sin 2 2 θ ) = = 4 cos 2 θ ( 2 + 2 cos θ ) = 8 cos 2 θ ( 1 + cos θ ) = = 16 cos 2 θ cos 2 θ 2 u^2+v^2=u=4\cos^2\theta(\cos\theta+\cos2\theta)^2+\\
+4\cos^2\theta(\sin\theta+\sin2\theta)^2=\\
=4\cos^2\theta(\cos^2\theta+2\cos\theta\cos2\theta+\cos^22\theta+\\
+\sin^2\theta+2\sin\theta\sin2\theta+\sin^22\theta)=\\
=4\cos^2\theta(2+2\cos\theta)=8\cos^2\theta(1+\cos\theta)=\\
=16\cos^2\theta\cos^2\frac{\theta}{2} u 2 + v 2 = u = 4 cos 2 θ ( cos θ + cos 2 θ ) 2 + + 4 cos 2 θ ( sin θ + sin 2 θ ) 2 = = 4 cos 2 θ ( cos 2 θ + 2 cos θ cos 2 θ + cos 2 2 θ + + sin 2 θ + 2 sin θ sin 2 θ + sin 2 2 θ ) = = 4 cos 2 θ ( 2 + 2 cos θ ) = 8 cos 2 θ ( 1 + cos θ ) = = 16 cos 2 θ cos 2 2 θ
#340153 on Dec 2023
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