soluciones_complejos

MATEMÁTICA II NÚMEROS COMPLEJOS

Soluciones Ejercicios Pág. 4. Tomo I 1. a.

(2 + i)

2

= 4 + 4i + i 2 = 3 + 4i

4

=

b.

(2 + i)

(( 2 + i ) )

2 2

= ( 3 + 4i ) = 9 + 24i + 16i 2 = −7 + 24i 2

c.

1 1( −i ) −i = = = −i i i (−i ) −i 2

d.

(1 − 3i )(3 + i ) = 3 + i − 9i − 3i 2 = 6 − 8i

e.

i (2 − i) i 2i − i 2 1 + 2i 1 2 = = = = + i 2 2 + i ( 2 + i )( 2 − i ) 4 − 2i + 2i − i 5 5 5

f. i 3 = i.i 2 = −i 1 − i 3 1 + i (1 + i )( 2 − i ) 2 − i + 2i − i 2 3 + i 3 1 = = = = = + i 2 + i 2 + i ( 2 + i )( 2 − i ) 4 − 2i + 2i − i 2 5 5 5

g. i 3 = −i i 4 = i 2 .i 2 = 1 i 7 = i 3 .i 4 = −i 1 − i + i 3 + 2i 4 + 5i 7 = 1 − i − i + 2 − 5i = 3 − 7i

h. i −12 =

1 1 = =1 12 2 6 i i ( )

1 1 1 − i 3 )( 5 + 3i12 ) = (1 + i ) 8 = 2 + 2i ( 4 4

2. a.

3i = 3π

2

b.

2i 3 = −2i = 2−π

2

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c.

1 = 10

d. −1 + i = 2 3π 4

e.

(1 − i )3 = (1 − i ) 2 (1 − i ) = (1 − 2i + i 2 ) (1 − i ) = −2i (1 − i ) = −2 − 2i = 8 −3π 4

f.

1 + i (1 + i )(1 + i ) 1 + i + i + i 2 2i = = = = i = 1π 2 1 − i (1 − i )(1 + i ) 1 + i − i − i 2 2

g.

1 + i 3 = 2π

3

h.

1

(1 − i )

2

=

( )

1 1 1(2i ) 2i 1 = = = = i= 1 2 2 2 1 − 2i + i −2i (−2i )(2i ) −4i 2

π

2

3. a. z = a 2 + b 2 = a 2 + ( −b ) = z 2

b. z.z = (a + bi )(a − bi ) = a 2 − abi + abi − b 2i 2 = a 2 + b 2 =

(

a 2 + b2

)

2

=( z)

2

c. z + z = (a + bi ) + (a − bi ) = a + bi + a − bi = 2a

d. z − z = (a + bi ) − (a − bi ) = a + bi − a + bi = 2bi

e. z + v = (a + bi ) + (c + di ) = ( a + c ) + (b + d )i = ( a + c ) − ( b + d ) i =

( a − bi ) + (c − di) = z + v f. z.v = ( a + bi ) (c + di ) = ac + adi + bci + bdi 2 = ( ac − bd ) + (ad + bc)i = ( ac − bd ) − (ad + bc)i  ⇒ z .v = (a − bi ) ( c − di ) = ac − adi − bci + bdi 2 = ( ac − bd ) − ( ad + bc ) i  z.v = z .v

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MATEMÁTICA II NÚMEROS COMPLEJOS

4. a. z = rϕ = r (cos ϕ + isenϕ )   ⇒ z.v = r (cos ϕ + isenϕ ) ρ (cos δ + isenδ ) = v = ρδ = ρ (cos δ + isenδ )  r ρ ( cos ϕ cos δ + i cos ϕ senδ + isenϕ cos δ + i 2 senϕ senδ ) =    r ρ ( cos ϕ cos δ − senϕ senδ ) + i ( cos ϕ senδ + senϕ cos δ )  = r ρ cos (ϕ + δ ) + isen (ϕ + δ ) 

  sen(ϕ +δ ) cos(ϕ +δ )   = ( r ρ )ϕ +δ

b. a = r cos ϕ z = a + bi = r ( cos ϕ + isenϕ ) = r cos ϕ + irsenϕ = rϕ ⇒  b = rsenϕ z = a − bi = r cos ϕ − irsenϕ   cos ϕ = cos ( −ϕ )  ⇒ z = r cos ( −ϕ ) + irsen ( −ϕ ) = r ( cos ( −ϕ ) + isen ( −ϕ ) ) = r−ϕ  − senϕ = sen ( −ϕ ) 

c. paso base k =2 z 2 = z.z = rϕ .rϕ =↑ ( r.r )ϕ +ϕ = r 2 2ϕ a)

Hip. Inductiva Supongo que z k −1 = ( r k −1 )( k −1)ϕ z k = z k −1.z = ( r k −1 )

( k −1)ϕ

.rϕ = ( r k −1.r )

( k −1)ϕ +ϕ

= ( r k −1+1 )

kϕ −ϕ +ϕ

= r k kϕ

5. z = 1ϕ = cos ϕ + isenϕ ⇒ z k = cos ( kϕ ) + isen ( kϕ ) ⊗ z 4 = ( cos ϕ + isenϕ ) = cos 4 ϕ + 4 cos 3 ϕ .isenϕ + 6 cos 2 ϕ . i 2 sen 2ϕ + 4 cos ϕ . i 3 sen3ϕ + i 4 sen 4ϕ 4

−1

−i

1

z 4 = ( cos 4 ϕ − 6 cos 2 ϕ sen 2ϕ + sen 4ϕ ) + i ( 4 cos3 ϕ senϕ − 4 cos ϕ sen3ϕ )  ⇒ ⇒ ⊗ z 4 = cos ( 4ϕ ) + isen ( 4ϕ )  cos ( 4ϕ ) = cos 4 ϕ − 6 cos 2 ϕ sen 2ϕ + sen 4ϕ sen ( 4ϕ ) = 4 cos3 ϕ senϕ − 4 cos ϕ sen3ϕ

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MATEMÁTICA II NÚMEROS COMPLEJOS

6.

z = 1ϕ = cos ϕ + isenϕ ⇒ z k = cos ( kϕ ) + isen ( kϕ ) ⊗ z 3 = ( cos ϕ + isenϕ ) = cos3 ϕ + 3cos 2 ϕ .isenϕ + 3cos ϕ .i 2 sen 2ϕ + i 3 sen3ϕ 3

z 3 = ( cos 3 ϕ − 3cos ϕ sen 2ϕ ) + i ( 3cos 2 ϕ .senϕ − sen3ϕ )  ⇒ ⇒ ⊗ z 3 = cos ( 3ϕ ) + isen ( 3ϕ )  cos ( 3ϕ ) = cos3 ϕ − 3cos ϕ sen 2ϕ sen ( 3ϕ ) = 3cos 2 ϕ .senϕ − sen3ϕ z = 1ϕ = cos ϕ + isenϕ ⇒ z k = cos ( kϕ ) + isen ( kϕ ) ⊗ z 2 = ( cos ϕ + isenϕ ) = cos 2 ϕ + 2 cos ϕ .isenϕ + i 2 sen 2ϕ 2

z 2 = ( cos 2 ϕ − sen 2ϕ ) + i ( 2 cos ϕ senϕ )  ⇒ ⇒ ⊗ z 2 = cos ( 2ϕ ) + isen ( 2ϕ )  cos ( 2ϕ ) = cos 2 ϕ − sen 2ϕ sen ( 2ϕ ) = 2 cos ϕ senϕ

7.   z = r−ϕ 3 3 4 z = rϕ ⇒  3 ⇒ z z = r . r = r 3ϕ −ϕ 2ϕ  3  3 4  z = r 3ϕ  ⇒ z z = i ⇔ r 2ϕ = 1π 2  i = 1π  2 r4 = 1⇒ r = 1 2ϕ = 2ϕ = 2ϕ =

π 2

π 2

π 2

⇒ϕ =π

4

⇒ z1 = 1π

+ 2π ⇒ ϕ = 5π + 4π ⇒ ϕ = 9π

4 4

4

= − 3π =π

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⇒ z2 = 1−3π

4

4

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MATEMÁTICA II NÚMEROS COMPLEJOS

8. z = a + bi ⇒ z = a − bi ⇒ z.z = (a + bi )(a − bi ) = a 2 − abi + abi − b 2i 2 = a 2 + b 2 = 4  ⇒ z 2 + z 2 = (a + bi ) 2 + (a − bi ) 2 = a 2 + 2abi + b 2i 2 + a 2 − 2abi + b 2i 2 = 2a 2 − 2b2 = −4  a 2 + b 2 = 4 (1)  2 2 a − b = −2 (2) a = 1 (1) + (2) ⇒ 2a 2 = 2 ⇒ a 2 = 1 ⇒   a = −1 b = 3 ⇒ z1 = 1 + 3i a = 1 ⇒ 1 + b2 = 4 ⇒ b2 = 3 ⇒  b = − 3 ⇒ z2 = 1 − 3i b = 3 ⇒ z3 = −1 + 3i a = −1 ⇒ 1 + b 2 = 4 ⇒ b 2 = 3 ⇒  b = − 3 ⇒ z4 = −1 − 3i

9. a. x 2 − 2 x + 17 = 0 ⇒ x =

2 ± −64 2 ± 64. −1 2 ± 8i 1 + 4i = = = 2 2 2 1 − 4i

b.

1 −1 4 −4 1 1

1

0

4

0

4

0

x + 4 = 0 ⇒ x = ± −4 = ± 4 −1 = ±2i 2

1  x = 2i −2i 

c.

x 4 + 16 = ( x 2 − 4i )( x 2 + 4i ) = 0

x 2 − 4i = 0 ⇒ x =

−0 ± 0 − 4.1( −4i ) 2

( ( 2 ) + isen (π 2 )) ⇒

i = cos π

i =i

1

2

=±

16i = ±2 i 2

2 2 π  π  = cos   + isen   = + i 2 4 4 2

x1 = 2 + 2i x2 = − 2 − 2i

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x 2 + 4i = 0 ⇒ x =

−0 ± 0 − 4.1( 4i ) 2

=±

−16i 16. −1. i =± = ±2i i 2 2

2 2 + i = − 2 + 2i 2 2  2 2  x4 = −2i  + i  = 2 − 2i 2 2   x3 = 2i

10. a.

(1 − 7i ) − 4 ( −4 + 28i ) = 1 − 14i + 49i 2 + 16 − 112i = −32 − 126i  ⇒ 2 ( 7 − 9i ) = 49 − 126i + 81i 2 = −32 − 126i  2 2 (1 − 7i ) − 4 ( −4 + 28i ) = ( 7 − 9i ) 2

b. z − (1 − 7i ) z + ( −4 + 28i ) = 0 ⇒ z = 2

(1 − 7i ) ± (1 − 7i ) 2

2

− 4 ( −4 + 28i )

=

 (1 − 7i ) + 7 − 9i = 4 − 8i = z1 (1 − 7i ) ± (7 − 9i) =  2  2  (1 − 7i ) − ( 7 − 9i ) = −3 + i = z 2  2 2

c.

z1 4 − 8i ( 4 − 8i )( −3 − i ) −12 − 4i + 24i + 8i 2 −20 + 20i = = = = = −2 + 2i z2 −3 + i ( −3 + i )( −3 − i ) 9 + 3i − 3i − i 2 10

 + 22 = 8  z1   ⇒ = 8 3π 4 z2  2  3π  ϕ = Arctg   =  −2  4  z2 −3 + i ( −3 + i ) (4 + 8i ) −12 − 24i + 4i + 8i 2 −20 − 20i 1 1 = = = = =− − i 2 z1 4 − 8i ( 4 − 8i ) (4 + 8i ) 16 + 32i − 32i − 64i 80 4 4 z1 = z2

( −2 )

2

2 2 z2 1   1  1  = −  +−  = z1 8  ⇒ z1 =  1   4  4  z2  8  −3π  3π 4 ϕ = Arctg (1) = −  4 

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MATEMÁTICA II NÚMEROS COMPLEJOS

11. a. z = 2 ⇒ a 2 + b2 = 2 ⇒ a 2 + b2 = 4

2

b.

2(a + bi ) − i = a + bi + i ⇒ 2a + (2b − 1)i = a + (b + 1)i ⇒ 4a 2 + (2b − 1) = a 2 + (b + 1) 2

2

4a 2 + 4b 2 − 4b + 1 = a 2 + b 2 + 2b + 1 ⇒ 3a 2 + 3b 2 − 6b = 0 ⇒ a 2 + b 2 − 2b = 0

c. a + bi + a − bi ≤

(

a 2 + b2

) ⇒ 2a ≤ a + b 2

2

2

⇒ a 2 + b 2 − 2a ≥ 0

C = (1, 0)

d. a + bi − i < a + bi + 1 ⇒ a + (b − 1)i < (a + 1) + bi ⇒ a 2 + (b − 1) 2 < ( a + 1) + b 2 ⇒ a 2 + b 2 − 2b + 1 < a 2 + 2a + 1 + b 2 ⇒ 2

2a + 2b > 0 ⇒ b > − a

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e. a + bi + i ≤ a + bi − i ⇒ a + (b + 1)i ≤ a + (b − 1)i ⇒ a 2 + (b + 1) 2 ≤ a 2 + (b − 1) 2 ⇒ b 2 + 2b + 1 ≤ b 2 − 2b + 1 ⇒ 4b ≤ 0 ⇒ b ≤ 0

f. a + bi + i ≤ 2 ( a + bi ) − 1 ⇒ a + ( b + 1) i ≤ ( 2a − 1) + 2bi ⇒ a 2 + ( b + 1) ≤ 2

( 2a − 1)

2

+ 4b 2

a 2 + b 2 + 2b + 1 ≤ 4a 2 − 4a + 1 + 4b 2 ⇒ 3a 2 + 3b 2 − 4a − 2b ≥ 0 ⇒ 4 2 a 2 + b2 − a − b ≥ 0 3 3  2 1 C = ,   3 3

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