Matemáticas I · Tema 1 · Sección 1.1

Ejercicios resueltos de operaciones elementales

4 ejercicios resueltos paso a paso del tema 1 de Matemáticas I (Fundamentos numéricos y números complejos). El enunciado está a la vista y la solución, plegada: intenta cada ejercicio antes de abrirla.

  • suma
  • resta
  • multiplicación
  • división
  • números complejos

Ejercicio 1

Dificultad: Básico

Dados z1=2+3iz_1 = 2 + 3i y z2=1−4iz_2 = 1 - 4i, calcular:

a) z1+z2z_1 + z_2

b) z1−z2z_1 - z_2

c) z1×z2z_1 \times z_2

d) z1÷z2z_1 \div z_2

Ver solución paso a paso4 pasos
  1. Paso 1
    Suma de números complejos

    z1+z2=(2+3i)+(1−4i)=(2+1)+(3−4)i=3−iz_1 + z_2 = (2 + 3i) + (1 - 4i) = (2 + 1) + (3 - 4)i = 3 - i

  2. Paso 2
    Resta de números complejos

    z1−z2=(2+3i)−(1−4i)=(2−1)+(3−(−4))i=1+7iz_1 - z_2 = (2 + 3i) - (1 - 4i) = (2 - 1) + (3 - (-4))i = 1 + 7i

  3. Paso 3
    Multiplicación

    z1×z2=(2+3i)(1−4i)=2⋅1+2⋅(−4i)+3i⋅1+3i⋅(−4i)z_1 \times z_2 = (2 + 3i)(1 - 4i) = 2 \cdot 1 + 2 \cdot (-4i) + 3i \cdot 1 + 3i \cdot (-4i)

    =2−8i+3i−12i2=2−5i+12=14−5i= 2 - 8i + 3i - 12i^2 = 2 - 5i + 12 = 14 - 5i

  4. Paso 4
    División usando el conjugado

    z1÷z2=2+3i1−4i⋅1+4i1+4i=(2+3i)(1+4i)(1−4i)(1+4i)z_1 \div z_2 = \frac{2 + 3i}{1 - 4i} \cdot \frac{1 + 4i}{1 + 4i} = \frac{(2 + 3i)(1 + 4i)}{(1 - 4i)(1 + 4i)}

    =2+8i+3i+12i21+16=2+11i−1217=−10+11i17=−1017+1117i= \frac{2 + 8i + 3i + 12i^2}{1 + 16} = \frac{2 + 11i - 12}{17} = \frac{-10 + 11i}{17} = -\frac{10}{17} + \frac{11}{17}i

Ejercicio 2

Dificultad: Intermedio

Dados z1=−1+2iz_1 = -1 + 2i, z2=3−iz_2 = 3 - i y z3=2+5iz_3 = 2 + 5i, calcular:

a) z1+z2+z3z_1 + z_2 + z_3

b) z1z2+z3z_1z_2 + z_3

c) (z1+z2)×z3(z_1 + z_2) \times z_3

d) z1z2z3\frac{z_1z_2}{z_3}

Ver solución paso a paso4 pasos
  1. Paso 1
    Suma de tres números complejos

    z1+z2+z3=(−1+2i)+(3−i)+(2+5i)=(−1+3+2)+(2−1+5)i=4+6iz_1 + z_2 + z_3 = (-1 + 2i) + (3 - i) + (2 + 5i) = (-1 + 3 + 2) + (2 - 1 + 5)i = 4 + 6i

  2. Paso 2
    Multiplicación seguida de suma

    z1z2=(−1+2i)(3−i)=−3+i+6i−2i2=−3+7i+2=−1+7iz_1z_2 = (-1 + 2i)(3 - i) = -3 + i + 6i - 2i^2 = -3 + 7i + 2 = -1 + 7i

    z1z2+z3=(−1+7i)+(2+5i)=1+12iz_1z_2 + z_3 = (-1 + 7i) + (2 + 5i) = 1 + 12i

  3. Paso 3
    Suma seguida de multiplicación

    (z1+z2)=(−1+2i)+(3−i)=2+i(z_1 + z_2) = (-1 + 2i) + (3 - i) = 2 + i

    (z1+z2)×z3=(2+i)(2+5i)=4+10i+2i+5i2=4+12i−5=−1+12i(z_1 + z_2) \times z_3 = (2 + i)(2 + 5i) = 4 + 10i + 2i + 5i^2 = 4 + 12i - 5 = -1 + 12i

  4. Paso 4
    División de productos

    z1z2z3=−1+7i2+5i⋅2−5i2−5i=(−1+7i)(2−5i)4+25\frac{z_1z_2}{z_3} = \frac{-1 + 7i}{2 + 5i} \cdot \frac{2 - 5i}{2 - 5i} = \frac{(-1 + 7i)(2 - 5i)}{4 + 25}

    =−2+5i+14i−35i229=−2+19i+3529=33+19i29=3329+1929i= \frac{-2 + 5i + 14i - 35i^2}{29} = \frac{-2 + 19i + 35}{29} = \frac{33 + 19i}{29} = \frac{33}{29} + \frac{19}{29}i

Ejercicio 3

Dificultad: Avanzado

Sean z1=1−3iz_1 = 1 - 3i, z2=−2+4iz_2 = -2 + 4i, z3=3+2iz_3 = 3 + 2i y z4=1−iz_4 = 1 - i. Calcular:

a) z1z2−z3z4z_1z_2 - z_3z_4

b) z1+z3z2−z4\frac{z_1 + z_3}{z_2 - z_4}

c) (z1z3)(z2z4)(z_1z_3)(z_2z_4)

d) ∣z1z2z3z4∣\left|\frac{z_1z_2}{z_3z_4}\right|

Ver solución paso a paso4 pasos
  1. Paso 1
    Productos individuales y resta

    z1z2=(1−3i)(−2+4i)=−2+4i+6i−12i2=−2+10i+12=10+10iz_1z_2 = (1 - 3i)(-2 + 4i) = -2 + 4i + 6i - 12i^2 = -2 + 10i + 12 = 10 + 10i

    z3z4=(3+2i)(1−i)=3−3i+2i−2i2=3−i+2=5−iz_3z_4 = (3 + 2i)(1 - i) = 3 - 3i + 2i - 2i^2 = 3 - i + 2 = 5 - i

    z1z2−z3z4=(10+10i)−(5−i)=5+11iz_1z_2 - z_3z_4 = (10 + 10i) - (5 - i) = 5 + 11i

  2. Paso 2
    Suma y resta en numerador y denominador

    z1+z3=(1−3i)+(3+2i)=4−iz_1 + z_3 = (1 - 3i) + (3 + 2i) = 4 - i

    z2−z4=(−2+4i)−(1−i)=−3+5iz_2 - z_4 = (-2 + 4i) - (1 - i) = -3 + 5i

    z1+z3z2−z4=4−i−3+5i⋅−3−5i−3−5i=(4−i)(−3−5i)9+25=−12−20i+3i+5i234=−17−17i34=−12−12i\frac{z_1 + z_3}{z_2 - z_4} = \frac{4 - i}{-3 + 5i} \cdot \frac{-3 - 5i}{-3 - 5i} = \frac{(4 - i)(-3 - 5i)}{9 + 25} = \frac{-12 - 20i + 3i + 5i^2}{34} = \frac{-17 - 17i}{34} = -\frac{1}{2} - \frac{1}{2}i

  3. Paso 3
    Producto de productos

    z1z3=(1−3i)(3+2i)=3+2i−9i−6i2=3−7i+6=9−7iz_1z_3 = (1 - 3i)(3 + 2i) = 3 + 2i - 9i - 6i^2 = 3 - 7i + 6 = 9 - 7i

    z2z4=(−2+4i)(1−i)=−2+2i+4i−4i2=−2+6i+4=2+6iz_2z_4 = (-2 + 4i)(1 - i) = -2 + 2i + 4i - 4i^2 = -2 + 6i + 4 = 2 + 6i

    (z1z3)(z2z4)=(9−7i)(2+6i)=18+54i−14i−42i2=18+40i+42=60+40i(z_1z_3)(z_2z_4) = (9 - 7i)(2 + 6i) = 18 + 54i - 14i - 42i^2 = 18 + 40i + 42 = 60 + 40i

  4. Paso 4
    Módulo de cociente

    z1z2z3z4=10+10i5−i⋅5+i5+i=(10+10i)(5+i)25+1=50+10i+50i+10i226=40+60i26=20+30i13\frac{z_1z_2}{z_3z_4} = \frac{10 + 10i}{5 - i} \cdot \frac{5 + i}{5 + i} = \frac{(10 + 10i)(5 + i)}{25 + 1} = \frac{50 + 10i + 50i + 10i^2}{26} = \frac{40 + 60i}{26} = \frac{20 + 30i}{13}

    ∣z1z2z3z4∣=∣20+30i13∣=202+30213=400+90013=130013=101313\left|\frac{z_1z_2}{z_3z_4}\right| = \left|\frac{20 + 30i}{13}\right| = \frac{\sqrt{20^2 + 30^2}}{13} = \frac{\sqrt{400 + 900}}{13} = \frac{\sqrt{1300}}{13} = \frac{10\sqrt{13}}{13}

Ejercicio 4

Dificultad: Experto

Dados los números complejos w=2−iw = 2 - i, u=−1+3iu = -1 + 3i, v=4+2iv = 4 + 2i y s=3−4is = 3 - 4i, demostrar que:

a) (wu)(vs)=(wv)(us)(wu)(vs) = (wv)(us)

b) ∣wuv∣=∣w∣∣u∣∣v∣|wuv| = |w||u||v|

c) wu+vs‾=wu‾+vs‾\overline{wu + vs} = \overline{wu} + \overline{vs}

d) Calcular wuvs2\frac{wuv}{s^2} y expresar en forma a+bia + bi

Ver solución paso a paso4 pasos
  1. Paso 1
    Verificación de la propiedad asociativa del producto

    wu=(2−i)(−1+3i)=−2+6i+i−3i2=−2+7i+3=1+7iwu = (2 - i)(-1 + 3i) = -2 + 6i + i - 3i^2 = -2 + 7i + 3 = 1 + 7i

    vs=(4+2i)(3−4i)=12−16i+6i−8i2=12−10i+8=20−10ivs = (4 + 2i)(3 - 4i) = 12 - 16i + 6i - 8i^2 = 12 - 10i + 8 = 20 - 10i

    (wu)(vs)=(1+7i)(20−10i)=20−10i+140i−70i2=20+130i+70=90+130i(wu)(vs) = (1 + 7i)(20 - 10i) = 20 - 10i + 140i - 70i^2 = 20 + 130i + 70 = 90 + 130i

    wv=(2−i)(4+2i)=8+4i−4i−2i2=8+2=10wv = (2 - i)(4 + 2i) = 8 + 4i - 4i - 2i^2 = 8 + 2 = 10

    us=(−1+3i)(3−4i)=−3+4i+9i−12i2=−3+13i+12=9+13ius = (-1 + 3i)(3 - 4i) = -3 + 4i + 9i - 12i^2 = -3 + 13i + 12 = 9 + 13i

    (wv)(us)=10(9+13i)=90+130i(wv)(us) = 10(9 + 13i) = 90 + 130i ✓

  2. Paso 2
    Verificación de la propiedad del módulo del producto

    wuv=(1+7i)(4+2i)=4+2i+28i+14i2=4+30i−14=−10+30iwuv = (1 + 7i)(4 + 2i) = 4 + 2i + 28i + 14i^2 = 4 + 30i - 14 = -10 + 30i

    ∣wuv∣=∣−10+30i∣=100+900=1000=1010|wuv| = |-10 + 30i| = \sqrt{100 + 900} = \sqrt{1000} = 10\sqrt{10}

    ∣w∣∣u∣∣v∣=∣2−i∣∣−1+3i∣∣4+2i∣=5⋅10⋅20=1000=1010|w||u||v| = |2 - i||-1 + 3i||4 + 2i| = \sqrt{5} \cdot \sqrt{10} \cdot \sqrt{20} = \sqrt{1000} = 10\sqrt{10} ✓

  3. Paso 3
    Verificación de la propiedad del conjugado de la suma

    wu+vs=(1+7i)+(20−10i)=21−3iwu + vs = (1 + 7i) + (20 - 10i) = 21 - 3i

    wu+vs‾=21−3i‾=21+3i\overline{wu + vs} = \overline{21 - 3i} = 21 + 3i

    wu‾+vs‾=1+7i‾+20−10i‾=(1−7i)+(20+10i)=21+3i\overline{wu} + \overline{vs} = \overline{1 + 7i} + \overline{20 - 10i} = (1 - 7i) + (20 + 10i) = 21 + 3i ✓

  4. Paso 4
    Cálculo del cociente complejo

    s2=(3−4i)2=9−24i+16i2=9−24i−16=−7−24is^2 = (3 - 4i)^2 = 9 - 24i + 16i^2 = 9 - 24i - 16 = -7 - 24i

    wuvs2=−10+30i−7−24i⋅−7+24i−7+24i=(−10+30i)(−7+24i)49+576\frac{wuv}{s^2} = \frac{-10 + 30i}{-7 - 24i} \cdot \frac{-7 + 24i}{-7 + 24i} = \frac{(-10 + 30i)(-7 + 24i)}{49 + 576}

    =70−240i−210i+720i2625=70−450i−720625=−650−450i625=−2625−1825i= \frac{70 - 240i - 210i + 720i^2}{625} = \frac{70 - 450i - 720}{625} = \frac{-650 - 450i}{625} = -\frac{26}{25} - \frac{18}{25}i