Resuelve las divisiones de números complejos y exprese el resultado en la forma: $a+bi$.

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Luis Hernandez

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Luis Hernandez 6 years, 1 month ago

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Luis Hernandez 6 years, 1 month ago

Created this as a copy of Arithmetics of complex numbers III.

Name	Status	Author	Last Modified
Arithmetics of complex numbers III	Ready to use	Newcastle University Mathematics and Statistics	25/05/2022 21:30
Complex Numbers Ex Sheet 3	draft	Violeta CIT	17/10/2017 19:54
Complex Numbers Ex Sheet 3	draft	Violeta CIT	17/10/2017 19:55
John's copy of Complex Numbers Ex Sheet 3	draft	John Sheekey	24/01/2018 12:30
Harry's copy of John's copy of Complex Numbers Ex Sheet 3	draft	Harry Flynn	11/06/2018 15:32
Luis's copy of Arithmetics of complex numbers III	draft	Luis Hernandez	30/11/2018 17:29
Resuelve las divisiones de números complejos y exprese el resultado en la forma: $a+bi$ .	Ready to use	Luis Hernandez	08/11/2020 01:17
Maria's copy of Arithmetics of complex numbers III	draft	Maria Aneiros	25/05/2019 08:20
Paula's copy of Arithmetics of complex numbers III	draft	Paula Ruiz-Hincapie	05/09/2019 13:33
Inverse and division of complex numbers	draft	Shaheen Charlwood	19/02/2020 14:24
Rechnen mit komplexen Zahlen III	Ready to use	Ulrich Görtz	08/02/2021 19:54
Dividing with complex numbers	Ready to use	Shaheen Charlwood	27/10/2022 18:53
Divisão de números complexos I	Ready to use	Isabel Sousa	30/04/2023 12:21

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Question statement

Resuelve las divisiones de números complejos y exprese el resultado en la forma: $a+bi$ .

Ingrese $a$ y $b$ como fracciones o enteros y no como decimales.

Give any introductory information the student needs.

			Name	Type	Generated Value

s3	integer	1
s2	integer	-1
s1	integer	1
s4	integer	-1
a3	integer	1
rz3	integer	1
c2	integer	2
c1	integer	7
z1	number	-9 + 3i
z2	number	-10 - 6i
z3	number	1 + 5i

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Part a) Gap-fill
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Part b) Gap-fill
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Part c) Gap-fill
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Part d) Gap-fill
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$\displaystyle \simplify[std]{{c1}/{z1}}\;=\;$ Unnamed gap.

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Advice

La división de dos números complejos se puede realizar multiplicando el numerador y el denominador por el conjugado del denominador.
Suponer que

$z = \frac{a+bi}{c+di},\;\; c+di \neq 0$ entonces tenemos:

$\begin{eqnarray*} z&=&\frac{a+bi}{c+di}\\ &=&\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\ &=&\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}\\ &=&\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i \end{eqnarray*}$

Aunque esta es una fórmula para el inverso, la mejor manera de encontrar estos números complejos es recordar multiplicar la parte superior e inferior por el conjugado del denominador.

(a)

$\begin{eqnarray*}\simplify[std]{{c1}/{z1}} &=&\simplify[std]{({c1}*{conj(z1)})/({z1}*{conj(z1)})}\\ &=&\simplify[std]{{c1*conj(z1)}/{abs(z1)^2}}\\ &=& \simplify[std]{{c1*re(z1)}/{abs(z1)^2}-{c1*im(z1)}/{abs(z1)^2}*i} \end{eqnarray*}$
(b)

$\begin{eqnarray*}\simplify[std]{{c2}/{z2}} &=&\simplify[std]{({c2}*{conj(z2)})/({z2}*{conj(z2)})}\\ &=&\simplify[std]{{c2*conj(z2)}/{abs(z2)^2}}\\ &=& \simplify[std]{{c2*re(z2)}/{abs(z2)^2}-{c2*im(z2)}/{abs(z2)^2}*i} \end{eqnarray*}$
(c)

$\begin{eqnarray*}\simplify[std]{{z1}/{z3}} &=&\simplify[std]{({z1}*{conj(z3)})/({z3}*{conj(z3)})}\\ &=&\simplify[std]{{z1*conj(z3)}/{abs(z3)^2}}\\ &=& \simplify[std]{{re(z1*conj(z3))}/{abs(z3)^2}+{im(z1*conj(z3))}/{abs(z3)^2}*i} \end{eqnarray*}$
(d)

$\begin{eqnarray*}\simplify[std]{{z3}/{z2}} &=&\simplify[std]{({z3}*{conj(z2)})/({z2}*{conj(z2)})}\\ &=&\simplify[std]{{z3*conj(z2)}/{abs(z2)^2}}\\ &=& \simplify[std]{{re(z3*conj(z2))}/{abs(z2)^2}+{im(z3*conj(z2))}/{abs(z2)^2}*i} \end{eqnarray*}$

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Resuelve las divisiones de números complejos y exprese el resultado en la forma: $a+bi$ .

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Luis Hernandez 6 years, 1 month ago

Luis Hernandez 6 years, 1 month ago

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Resuelve las divisiones de números complejos y exprese el resultado en la forma: a+bia+bi.

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Luis Hernandez 6 years, 1 month ago

Luis Hernandez 6 years, 1 month ago

Error in variable testing condition

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Generated value: integer

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Resuelve las divisiones de números complejos y exprese el resultado en la forma: $a+bi$ .

Generated value: `integer`