Haynes Nunez
04/29/2023 · Primary School
Use la forma polar \( d \) y úna cada raíz con \( u \) a) \( z^{6}=64 \) b) \( z^{6}=i \) c) \( z^{5}-32=0 \) d) \( z^{4}=1+i \)
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a) \( z^{6}=64 \)
Primero, convertimos 64 a forma polar:
\[ 64 = 2^6 \]
\[ r = 2^6 = 64 \]
\[ \theta = 0 \] (porque está en el eje real positivo)
Entonces, \( z = 64(\cos(0) + i\sin(0)) \).
Para encontrar las raíces, dividimos \( \theta \) por 6:
\[ \theta_k = \frac{0}{6} + \frac{2\pi k}{6} = \frac{\pi k}{3} \]
Las raíces son:
\[ z_k = 64(\cos(\frac{\pi k}{3}) + i\sin(\frac{\pi k}{3})) \]
para \( k = 0, 1, 2, 3, 4, 5 \).
b) \( z^{6}=i \)
Primero, convertimos \( i \) a forma polar:
\[ i = 0 + i \]
\[ r = 1 \]
\[ \theta = \frac{\pi}{2} \]
Entonces, \( z = 1(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) \).
Para encontrar las raíces, dividimos \( \theta \) por 6:
\[ \theta_k = \frac{\pi}{2} + \frac{2\pi k}{6} = \frac{\pi}{2} + \frac{\pi k}{3} \]
Las raíces son:
\[ z_k = 1(\cos(\frac{\pi}{2} + \frac{\pi k}{3}) + i\sin(\frac{\pi}{2} + \frac{\pi k}{3})) \]
para \( k = 0, 1, 2, 3, 4, 5 \).
c) \( z^{5}-32=0 \)
Primero, convertimos 32 a forma polar:
\[ 32 = 2^5 \]
\[ r = 2^5 = 32 \]
\[ \theta = 0 \] (porque está en el eje real positivo)
Entonces, \( z = 32(\cos(0) + i\sin(0)) \).
Para encontrar las raíces, dividimos \( \theta \) por 5:
\[ \theta_k = \frac{0}{5} + \frac{2\pi k}{5} = \frac{2\pi k}{5} \]
Las raíces son:
\[ z_k = 32(\cos(\frac{2\pi k}{5}) + i\sin(\frac{2\pi k}{5})) \]
para \( k = 0, 1, 2, 3, 4 \).
d) \( z^{4}=1+i \)
Primero, convertimos \( 1+i \) a forma polar:
\[ 1+i = \sqrt{1^2 + 1^2}(\cos(\arctan(1)) + i\sin(\arctan(1))) \]
\[ 1+i = \sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})) \]
\[ 1+i = \sqrt{2}(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}) \]
\[ 1+i = 1 + i \]
Entonces, \( r = \sqrt{2} \)
\[ \theta = \frac{\pi}{4} \]
Para encontrar las raíces, dividimos \( \theta \) por 4:
\[ \theta_k = \frac{\pi}{4} + \frac{2\pi k}{4} = \frac{\pi}{4} + \frac{\pi k}{2} \]
Las raíces son:
\[ z_k = \sqrt{2}(\cos(\frac{\pi}{4} + \frac{\pi k}{2}) + i\sin(\frac{\pi}{4} + \frac{\pi k}{2})) \]
para \( k = 0, 1, 2, 3 \).
Quick Answer
a) \( z^{6}=64 \) raíces: \( z_k = 64(\cos(\frac{\pi k}{3}) + i\sin(\frac{\pi k}{3})) \) para \( k = 0, 1, 2, 3, 4, 5 \).
b) \( z^{6}=i \) raíces: \( z_k = 1(\cos(\frac{\pi}{2} + \frac{\pi k}{3}) + i\sin(\frac{\pi}{2} + \frac{\pi k}{3})) \) para \( k = 0, 1, 2, 3, 4, 5 \).
c) \( z^{5}-32=0 \) raíces: \( z_k = 32(\cos(\frac{2\pi k}{5}) + i\sin(\frac{2\pi k}{5})) \) para \( k = 0, 1, 2, 3, 4 \).
d) \( z^{4}=1+i \) raíces: \( z_k = \sqrt{2}(\cos(\frac{\pi}{4} + \frac{\pi k}{2}) + i\sin(\frac{\pi}{4} + \frac{\pi k}{2})) \) para \( k = 0, 1, 2, 3 \).
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