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Transcript of Ims
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Introducción a la Matemática Superior
8. Demostrar que ¿∨z−w∨¿2≤(1+||z||2)
||z−w||2=( z−w ) ( z−w )
¿|z−w|∨¿2=z . z−w z−z w+w .w ¿
¿|z−w|∨¿2=¿|z|∨¿2−2ℜ ( z w )+¿|w|∨¿2 ¿¿¿
¿|z−w|∨¿2=¿|z|∨¿2+2||z||.||w||+¿|w|∨¿2≤1+||z||2+||w||2+||z||2 .¿|w|∨¿2¿¿¿¿
¿|z−w|∨¿2≤1+||z||2+||w||2+||z||2 .∨|w|∨¿¿
¿|z−w|∨¿2≤(1+||z||2)+(1+||w||2)¿
12. Resolver en los complejos: ||z+2 i||=¿|iz−2|∨¿
||a−bi+2 i||=¿|−b−2+ai|∨¿
||a+(2−b) i||=√ (− (b+2 )2 )+a2
a2+4−4 b+b2=b2+4b+4+a2
−4b=4b
0=8b→b=0
z=a+bi
z=a+0 i
z=acte
z∈ R
13.
a)R={z∈C /¿|z+1|∨≤4−¿|z−1|∨¿}Z=x+ yi
||z+1||≤k−¿|z−1|∨¿
||(x+1 )+ yi||≤4−¿|(x−1 )+ yi|∨¿
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√¿¿
( x+1 )2+ y2≤16−8√( x−1 )2+ y2+( x−1 )2+ y2
(x+1)2−( x−1 )2≤16−8√(x−1)2+ y2
x≤ 4−2√(x−1)2+ y2
2√(x−1)2+ y2≤4−x
4 [ ( x−1 )2+ y2 ]≤16−8 x+x2
4 (x2−2x+1 )+4 y2≤16−8 x+x2
4 x2−8 x+4+4 y2≤16−8 x+x2
3 x2+4 y2≤12
x2
4+ y2
3≤1
b) R={z∈C /||z+1−2 i||+||iz+2−3 i||≤6 }
z=a+bi
||z+1−2i||+||iz+2−3i||≤6……I
z ¿
iz+2−3 i=i (a+bi )+2−3 i=ia−b+2−3 i=(2−b )+(a−3 ) i
En I
¿
||(a+1 )+(b−2 ) i||≤6−¿|(2−b )+(a−3 ) i|∨¿
elevandoal cuadrado
(a+1)2+(b−2)2≤36−12√(2−b )2+(a−3 )2+(2−b )2+(a−3)2
(a+1)2− (a−3 )2+¿
a2+2a+1¿−(a2−6a+9 )≤36−12√ (2−b )2+(a−3)2
8a−8≤36−12√(2−b)2+(a−3)2
2a−2≤9−3√(2−b)2+(a−3)2
3√(2−b)2+(a−3)2≤11−2a
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elevandoal cuadrado
9¿
9 [ (4−4 b+b2 )+(a2−6 a+9 ) ]≤121−44a+4 a2
9 [b2−4 b+a2−6a+13 ]≤121−44a+4a2
9b2−36b+9 a2−54 a+117≤121−44a+4a2
9b2−36b+5a2−10a−4≤0
9b2−36b+5a2−10a≤4
9 (b2−4 b+4 )−36+5 (a2−2a+1 )−5≤4
9 (b−2 )2+5(a−1)2≤45
(b−2 )2
5+(a−1)2
9≤1
c) R={z∈C /ℑ ( z2 )=9 }
z=a+bi→ z2=a2+2abi−b
a2=b
2ab=9
ab=92→a3=9
2→a= 3√9 /2→a=1.65
Reemplazando: Z=3√9/2+ 3√9/22
e) M={ zϵ C /0<ℜ(iz)≤1 }
z=a+i
zi=−b+ai
zi=−bℜ +ai
0←b≤1
−1<b≤0
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f) S={z ϵ C /2 z z+ (2+ i) z+(2−i ) z=2}
2(x + yi)(x – yi) + (2 + i)(x + yi) + (2 – i)(x – yi) = 2
2(x2 –y2i2) + 2x + 2yi + xi + yi2 +2x-2yi – xi + yi2 = 2
2(x2 + y2) +4x + 2yi2 = 2
X2 +y2 + 2x – y = 1
(X+1)2 -1+ (y- ½)2 – ¼ = 1
(X+1)2 + (y- ½)2 = 9/4
H= -1 k= ½ R= 3/2
18. Reducir6cis30°∗2cis10°3cis(−20 °)
4 (cos 30°+isen30 ° ) (cos 10°+isen10 ° )cos 20°−isen20 °
4 e(30 i )e (10 i)
e (−20i )
4 e(60 i)
4 ¿)
4 ( 12+ √32
i)
(2+2√3i)
2(1+√3 i)
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31. Resolver las siguientes ecuaciones:
c) z6+7 z3−¿8=0z6+z3−z3+7 z3−8=0
z3 ( z3−1 )+8 ( z3−1 )=0
(z¿¿3+8) ( z3−1 )=0¿
( z+2 ) ( z2−2 z+4 ) ( z−1 ) ( z2+z+1 )=0
z=−2 z=1±√3 i z=1 z=−1±√3 i2