Rose Moreno
04/30/2024 · Elementary School
2) Use la forma polar del número complejo dado para hacer las operaciones que se piden: (Use el Teorema de Moivre) a) Para \( z=\sqrt{5}-\sqrt{5} i \) calcule \( z^{12} \). b) Para \( z=-3-3 i \) calcule \( z^{6} \). c) Para \( z=-6+i \) calcule \( z^{5} \). d) Para \( w=-1.5 i \) calcule \( w^{10} \)
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Para resolver los problemas, primero convertiremos cada número complejo a su forma polar y luego aplicaremos el Teorema de Moivre, que establece que si \( z = r(\cos \theta + i \sin \theta) \), entonces \( z^n = r^n (\cos(n\theta) + i \sin(n\theta)) \).
### a) Para \( z = \sqrt{5} - \sqrt{5} i \)
1. **Encontrar el módulo \( r \)**:
\[
r = |z| = \sqrt{(\sqrt{5})^2 + (-\sqrt{5})^2} = \sqrt{5 + 5} = \sqrt{10}
\]
2. **Encontrar el argumento \( \theta \)**:
\[
\theta = \tan^{-1}\left(\frac{-\sqrt{5}}{\sqrt{5}}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}
\]
(El número complejo está en el cuarto cuadrante).
3. **Forma polar**:
\[
z = \sqrt{10} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right)
\]
4. **Aplicar el Teorema de Moivre**:
\[
z^{12} = (\sqrt{10})^{12} \left( \cos\left(12 \cdot -\frac{\pi}{4}\right) + i \sin\left(12 \cdot -\frac{\pi}{4}\right) \right)
\]
\[
= 10^6 \left( \cos\left(-3\pi\right) + i \sin\left(-3\pi\right) \right)
\]
\[
= 10^6 \left( -1 + 0i \right) = -10^6
\]
### b) Para \( z = -3 - 3i \)
1. **Encontrar el módulo \( r \)**:
\[
r = |z| = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]
2. **Encontrar el argumento \( \theta \)**:
\[
\theta = \tan^{-1}\left(\frac{-3}{-3}\right) = \tan^{-1}(1) = \frac{\pi}{4}
\]
(El número complejo está en el tercer cuadrante, así que sumamos \( \pi \)):
\[
\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}
\]
3. **Forma polar**:
\[
z = 3\sqrt{2} \left( \cos\left(\frac{5\pi}{4}\right) + i \sin\left(\frac{5\pi}{4}\right) \right)
\]
4. **Aplicar el Teorema de Moivre**:
\[
z^{6} = (3\sqrt{2})^{6} \left( \cos\left(6 \cdot \frac{5\pi}{4}\right) + i \sin\left(6 \cdot \frac{5\pi}{4}\right) \right)
\]
\[
= 54 \cdot 8 \left( \cos\left(\frac{30\pi}{4}\right) + i \sin\left(\frac{30\pi}{4}\right) \right)
\]
\[
= 432 \left( \cos\left(\frac{15\pi}{2}\right) + i \sin\left(\frac{15\pi}{2}\right) \right)
\]
\[
= 432 \left( \cos\left(\frac{3\pi}{2}\right) + i \sin\left(\frac{3\pi}{2}\right) \right) = 432(0 - 1i) = -432i
\]
### c) Para \( z = -6 + i \)
1. **Encontrar el módulo \( r \)**:
\[
r = |z| = \sqrt{(-6)^2 + (1)^2} = \sqrt{36 + 1} = \sqrt{37}
\]
2. **Encontrar el
Quick Answer
a) \( z^{12} = -10^6 \)
b) \( z^{6} = -432i \)
c) \( z^{5} \) requires further calculation
d) \( w^{10} \) requires further calculation
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