Question

porfavoooor ayudaaa de urgenciaa

Answer 1

Los números complejos resultantes de las operaciones son:

$35+\frac{216}{5}i$

$-\frac{199}{5}+63i$

$\frac{108}{25}-\frac{72}{5}i$

$-\frac{96}{205}-\frac{7}{82}i$

Explicación paso a paso:

1) En esta operación hacemos primero la resta entre paréntesis y luego el producto:

$Z_1(Z_2-Z_3)=(-\frac{6}{5}+4i)((\frac{7}{2}-6i)-(-4+5i))\\\\Z_1(Z_2-Z_3)=(-\frac{6}{5}+4i)(\frac{15}{2}-11i)\\\\Z_1(Z_2-Z_3)=-\frac{6}{5}.\frac{15}{2}+(-\frac{6}{5})(-11i)+4i.\frac{15}{2}+4i(-11i)\\\\Z_1(Z_2-Z_3)=-9+\frac{66}{5}i+30i+44\\\\Z_1(Z_2-Z_3)=35+\frac{216}{5}i$

2) En este punto podemos eliminar el paréntesis distribuyendo el signo (-) a los dos términos dentro de él:

$4Z_1-(2Z_2-7Z_3)=4Z_1-2Z_2+7Z_3\\\\4Z_1-(2Z_2-7Z_3)=4(-\frac{6}{5}+4i)-2(\frac{7}{2}-6i)+7(-4+5i)\\\\4Z_1-(2Z_2-7Z_3)=-\frac{24}{5}+16i-(7-12i)+(-28+35i)\\\\4Z_1-(2Z_2-7Z_3)=-\frac{199}{5}+63i$

3) Aquí vamos a realizar primero la resta entre paréntesis y luego el producto por Z1:

$\frac{4}{5}Z_1(Z_2-2Z_3)=\frac{4}{5}(-\frac{6}{5}+4i)((\frac{7}{2}-6i)-2(-4+5i))\\\\\frac{4}{5}Z_1(Z_2-2Z_3)=\frac{4}{5}(-\frac{6}{5}+4i)(\frac{7}{2}-6i+8-10i)\\\\\frac{4}{5}Z_1(Z_2-2Z_3)=\frac{4}{5}(-\frac{6}{5}+4i)(\frac{23}{2}-16i)\\\\\frac{4}{5}Z_1(Z_2-2Z_3)=\frac{4}{5}((-\frac{6}{5})\frac{23}{2}+(-\frac{6}{5})(-16i)+4i(\frac{23}{2})+4i(-16i))\\\\\frac{4}{5}Z_1(Z_2-2Z_3)=\frac{4}{5}(-\frac{69}{5}+\frac{96}{5}i+46i-64i)$

Ahora, distribuyendo la fracción queda:

$\frac{4}{5}Z_1(Z_2-2Z_3)=\frac{4}{5}(\frac{27}{5}i-18i)\\\\\frac{4}{5}Z_1(Z_2-2Z_3)=\frac{108}{25}i-\frac{72}{5}i$

4) Primero hacemos la suma en el numerador:

$\frac{Z_1+Z_2}{Z_3}=\frac{-\frac{6}{5}+4i+\frac{7}{2}-6i}{-4+5i}\\\\\frac{Z_1+Z_2}{Z_3}=\frac{\frac{23}{10}-2i}{-4+5i}$

Para hacer la división multiplicamos y dividimos por el conjugado del denominador:

$\frac{Z_1+Z_2}{Z_3}=\frac{\frac{23}{10}-2i}{-4+5i}\frac{-4-5i}{-4-5i}=\frac{\frac{23}{10}(-4)-5i(\frac{23}{10})+(-2i)(-4)+(-2i)(-5i)}{(-4)(-4)+(-4)(-5i)+5i(-4)+5i(-5i)}\\\\\frac{Z_1+Z_2}{Z_3}=\frac{\frac{23}{10}-2i}{-4+5i}\frac{-4-5i}{-4-5i}=\frac{-\frac{46}{5}-(\frac{23}{2})i+8i-10}{16+25}\\\\\frac{Z_1+Z_2}{Z_3}=\frac{\frac{23}{10}-2i}{-4+5i}\frac{-4-5i}{-4-5i}=\frac{-\frac{96}{5}-(\frac{7}{2})i}{41}=-\frac{96}{205}-\frac{7}{82}i$