Park Carter
10/04/2023 · Senior High School

Use la forma polar d a) Para \( z=3-0.5 \) b) Para \( z=-64 i y \) c) Para \( z=-3-2 \) d) Para \( z=4-4 i \)

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Para convertir un número complejo a su forma polar, utilizamos la fórmula: \[ z = r(\cos \theta + i \sin \theta) \] donde \( r \) es el módulo del número complejo y \( \theta \) es el argumento (ángulo) del número complejo. El módulo se calcula como: \[ r = \sqrt{x^2 + y^2} \] y el argumento se calcula como: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \] donde \( x \) es la parte real y \( y \) es la parte imaginaria del número complejo \( z = x + yi \). Ahora, vamos a calcular la forma polar para cada uno de los números complejos dados: ### a) Para \( z = 3 - 0.5i \) 1. **Parte real**: \( x = 3 \) 2. **Parte imaginaria**: \( y = -0.5 \) **Módulo**: \[ r = \sqrt{3^2 + (-0.5)^2} = \sqrt{9 + 0.25} = \sqrt{9.25} \approx 3.041 \] **Argumento**: \[ \theta = \tan^{-1}\left(\frac{-0.5}{3}\right) \approx \tan^{-1}(-0.1667) \approx -0.1651 \text{ rad} \] Por lo tanto, la forma polar es: \[ z = 3.041 \left( \cos(-0.1651) + i \sin(-0.1651) \right) \] ### b) Para \( z = -64i \) 1. **Parte real**: \( x = 0 \) 2. **Parte imaginaria**: \( y = -64 \) **Módulo**: \[ r = \sqrt{0^2 + (-64)^2} = \sqrt{4096} = 64 \] **Argumento**: \[ \theta = \tan^{-1}\left(\frac{-64}{0}\right) = -\frac{\pi}{2} \text{ rad} \quad (\text{en el eje negativo imaginario}) \] Por lo tanto, la forma polar es: \[ z = 64 \left( \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right) \right) \] ### c) Para \( z = -3 - 2i \) 1. **Parte real**: \( x = -3 \) 2. **Parte imaginaria**: \( y = -2 \) **Módulo**: \[ r = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.606 \] **Argumento**: \[ \theta = \tan^{-1}\left(\frac{-2}{-3}\right) = \tan^{-1}\left(\frac{2}{3}\right) \approx 0.588 \text{ rad} \] Dado que el número está en el tercer cuadrante, sumamos \( \pi \): \[ \theta \approx 0.588 + \pi \approx 3.730 \text{ rad} \] Por lo tanto, la forma polar es: \[ z = 3.606 \left( \cos(3.730) + i \sin(3.730) \right) \] ### d) Para \( z = 4 - 4i \) 1. **Parte real**: \( x = 4 \) 2. **Parte imaginaria**: \( y = -4 \) **Módulo**: \[ r = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.657 \] **Argumento**: \[ \theta = \tan^{-1}\left(\frac{-4}{4}\right) = \tan^{-1}(-1) = -\frac{\pi}{4} \text{ rad} \] Por lo tanto, la forma polar es: \[ z = 4\sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right) \] ### Resumen de las formas polares: a) \( z = 3.041 \left(

Quick Answer

a) \( z = 3.041 \left( \cos(-0.1651) + i \sin(-0.1651) \right) \) b) \( z = 64 \left( \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right) \right) \) c) \( z = 3.606 \left( \cos(3.730) + i \sin(3.730) \right) \) d) \( z = 4\sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right) \)
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