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7. Calcula la parte real de la siguiente función \( f|z|=z^{3} \), si, \( z=x+i y \) \( \begin{array}{llll}\text { a) } \mathfrak{R} \mid f(z)=x^{3}-3 x y^{2} & \text { b) } \mathfrak{R} \mid f(z)=3 x^{2} y-y^{3} & \text { c) } \mathfrak{R}|f| z \mid=x^{3}+3 x y^{2} & \text { d) } \mathfrak{R}|f| z \mid=y^{3}-3 x^{2} y\end{array} \)

Ask by Mathis Chambers. in Mexico
Mar 20,2025

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Tutor-Verified Answer

Answer

La parte real de la función \( f(z) = z^3 \) es \( x^3 - 3xy^2 \).

Solution

La función se define como \[ f(z) = z^3, \] donde \( z = x + iy \). Para encontrar la parte real, expandimos el cubo: \[ z^3 = (x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3. \] Calculamos cada término por separado: 1. \( x^3 \) ya es real. 2. \( 3x^2(iy) = 3ix^2y \) es un término imaginario. 3. \( 3x(iy)^2 \): Notamos que \((iy)^2 = i^2y^2 = -y^2\), de modo que: \[ 3x(iy)^2 = 3x(-y^2) = -3xy^2, \] es un término real. 4. \( (iy)^3 \): Notamos que \((iy)^3 = i^3y^3 = -iy^3\) es un término imaginario. Reuniendo términos reales e imaginarios obtenemos: \[ z^3 = \bigl(x^3 - 3xy^2 \bigr) + i\bigl(3x^2y - y^3 \bigr). \] Por lo tanto, la parte real de \( f(z) = z^3 \) es \[ \mathfrak{R} \, [f(z)] = x^3 - 3xy^2. \] La opción correcta es: \[ \textbf{a) } \mathfrak{R} \, [f(z)]= x^3 - 3xy^2. \]

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Para resolver esta función, primero expresamos \( z \) como \( z = x + i y \). Luego, calculamos \( f(z) = z^3 \): \[ z^3 = (x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3. \] Desarrollando cada término nos queda: \[ = x^3 + 3x^2(iy) - 3xy^2 - iy^3. \] Al agrupar las partes real e imaginaria, tenemos: \[ = (x^3 - 3xy^2) + i(3x^2y - y^3). \] Por lo tanto, la parte real de \( f(z) \) es \( \mathfrak{R}(f(z)) = x^3 - 3xy^2 \). Por lo tanto, la respuesta correcta es la opción a) \( \mathfrak{R}(f(z)) = x^3 - 3xy^2 \).

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