Pumas goyayaya

6.- EXPRESAR CADA UNO DE LOS NUMEROS COMPLEJOS QUE SE DAN A CONTINUACION EN FORMA POLAR .

SOLUCION

A) 15[pic] = 15[pic] = 15/45°

B) 5[pic] = 5[pic] = 5/-120°

C) -4[pic] = -4[pic] = -4/30°

D) -2[pic] = -2[pic] = 2/90°

E) 10[pic] = 10 [pic] = 10/-210°F) -18 [pic] = -18[pic] =

6.- EFECTUAR LA OPERACIÓN QUE SE INDICA

z = 3 – ј4 hallar zz*

( 3 – ј4 ) ( 3 – ј4 ) = 9 + ј12 - ј12 -16 ј
= 9 + (12-12)ј – 16 (-1)
= 9 + (0)ј + 16

= 9+16=25

z = 10 /-40° hallar zz*

x = 10 cos -40
x = 10 (0.766044443)
x = 7.66

y = 10 sen -40
y = 10 (-0.642787609)
y = - 6.42z = 7.66 - ј 6.42

= (7.66 - ј 6.42) (7.66 + ј 6.42)
= 58.67 + ј49.17 - ј49.17 – 41.21ј
= 58.67 + ( ј49.17 - ј49.17 ) – 41.21 (-1)
= 58.67 + (0)ј + 41.21

58.67+41.21= 99.95 = 100

z = 20 /53.1° hallar z + z*

x = 20 cos 53.1
x = 20 ( 0.600420225)
x = 12

y = 20 sen 53.1
y = 20 (0.799684658)
y = 15.99 = 16

z = 12 + ј16

= (12 + ј16) (12 - ј16)
= 144 - ј192 + ј192- 256ј
= 144 + (- ј192 + ј192)- 256 (-1)
= 144 + (0)ј +256

= 144+256= 400

2.5е - јπ/3 hallar zz*

z = 2.5е – ј 180/3 z = 2.5е – ј60

(2.5е – ј60) (2.5е – ј60)
= 6.25 + ј150 – ј150 - 3600ј
= 6.25 + (150-150)ј -3600 (-1)
= 6.25 + (0)ј + 3600

= 6.25 +3600 = 3606.25

z = 2+ ј8 hallar z - z*

(2+ ј8) (2- ј8)

= 4 - ј18 + ј18 - 64ј
= 4 + (- ј18 + ј18) – 64 (-1)
= 4 + (0)ј+64

= 4+64 =68

z = 10 – ј4 hallar z + z*

(10 – ј4 ) (10 + ј4)

= 100 + ј40 - ј40 - 16ј
= 100 + (ј40 - ј40) – 16 (-1)
= 100 + (0)ј +16

= 100 + 16 = 116

z = 95 /25° hallar z - z*

x = 95 cos 25
x = 95 (0.906307787)
x = 86.09

y = 95 sen 25
y = 95 ( 0.422618261)
y= 40.14

z = 86.09 + ј40.14

(86.09 + ј40.14) (86.09 - ј40.14)

= 7396 – ј3440 + ј3400 – 1600ј
= 7396+ (– ј3440 + ј3400) – 1600 (-1)
= 7396 + (0)ј + 1600

= 7396 + 1600 = 8996

8.-PASAR DE FORMA POLAR A BINOMICA CADA UNO DE LOS NUMEROS COMNPLEJOS QUE SE INDICAN

12.3 /30°

X= 12.3cos 30° 12.3
X= 12.3 (0.866025403) ) 30°
X= 10.65

Y= 12.3sen 30°
Y= 12.3 (0.5)
Y= 6.15

Z= 10.65 + ј6.15

53/160°

X= 53cos 160°
X= 53 (- 0.93969262) 53
X= - 49.8 )160°

Y= 53sen 160°
Y= 53(0.342020143)
Y= 18.1

Z= - 49.8 + ј18.1

25/- 45

X= 25cos - 45°
X= 25 (0.7071067819 -45(
X= 17.6
25

Y= 25sen - 45°
Y= 25 (- 0.707106781)
Y= - 17.6

Z= 17.6 – ј17.6

86/ - 115°

X= 86cos -115°
X= 86 (- 0.422618261)
X= - 36.3 -115° (

Y= 86sen -115°
Y= 86 (- 0.906307787)
Y= -77.9 = 78

Z= - 36.3 – ј78

0.05 / - 20°

X= 0.05cos – 20° -20° (
X= 0.05 (0.93969262)X=0.046 0.05

Y= 0.05sen – 20°
Y= 0.05 (- 0.342020143)
Y= - 0.0171

Z=0.046 – ј0.0171

0.003 /80°

X= 0.003cos 80° 0.003
X= 0.003 (0.173648177) ) 80°
X= 0.00052

Y= 0.003sen 80°
Y= 0.003 (0.984807753)
Y= 0.00295

Z= 0.00052 + ј0.00295

0.013 /260°

X= 0.0013cos 260°
X= 0.0013 (-0.173648177)
X= - 0.00225 0.013
) 260°

Y= 0.013sen 260°
Y= 0.0013 (-0.984807753)
Y= -0.0128

Z= - 0.00225 – ј0.0128

0.156 /-190°

-190° (
X= 0.156cos -190°
X= 0.156 (-0.984807753) 0.156
X= - 0.1536

Y= 0.156sen -190°
Y= 0.156 (0.173648177)
Y= 0.0270

Z= - 0.1536 + ј0.0270

9.- PASAR DE FORMA BINOMICA A POLAR CADA UNO DE LOS NUMEROS COMPLEJOS QUE SE INDICAN

a) - 12 + ј16

ѳ= arc tag = [pic] = [pic] = -1.333
ј16
ѳ= [pic] = (-1.333) = -53.12
- 12

Y=[pic]= [pic] = - 20

ѳ= 180-53.12= 126.8

z= -20/126.8°

b) 2 – ј4

ѳ= arc tag = [pic] = [pic] = -2 2
-ј4
ѳ= [pic] = (-2) = -63.4

Y=[pic] = [pic] = 4.47

Z= 4.47/-63.43°

c) -59- ј25

ѳ= arc tag = [pic] = [pic] = 0.423 -59

-ј25
ѳ= [pic] = (0.423) = 22.9 = 23

Y= [pic] = [pic] = 63.98 = 64

Z= 64/203°

d) 700+ј200

ј200
ѳ= arc tag = [pic] = [pic] =...