Question

Dado los números complejos z=2( 1- i) + 3(i-2) y w= 1/(1+2i) determinar:
Re(w^2)
Im(i/zw)
|z+w|

Answer 1

Veamos.

z = 2 - 2i + 3i - 2 = i

w =1/(1 + 2i) = (1 - 2i)/5

w^2 = (- 3 - 4i)/25; su parte real es - 3/25

z w = (2 + i)/5

i/(z w) = 1 + 2i; su parte imaginaria es 2

z + w = (1 + 3i)/5

|z + w| = √10 / 5

He omitido las operaciones elementales. Revisa por si hay errores

Saludos Herminio

Answer 2

$z=2-2i+3i-6\to \boxed{z=-4+i} \\ \\ w=\dfrac{1}{1+2i}\\ \\ w=\dfrac{1-2i}{(1+2i)(1-2i)}\\ \\ \\ w=\dfrac{1-2i}{5}\\ \\ \boxed{w=\dfrac{1}{5}-\dfrac{2}{5}i}$

$\textcircled{1}\\ \\ w^2 =\left(\dfrac{1}{5}-\dfrac{2}{5}i\right)^2\\ \\ w^2=\dfrac{1}{25}-\dfrac{4}{25}i-\dfrac{4}{25}\\ \\ w^2=-\dfrac{3}{25}-\dfrac{4}{25}i\\ \\ \boxed{\mbox{Re}(w^2)=-\dfrac{3}{25}}$

$\textcircled{2}\\ \\ \dfrac{i}{zw}=\dfrac{i}{-4+i}\cdot (1+2i)\\ \\ \dfrac{i}{zw}=\dfrac{i(4+i)}{-17}\cdot (1+2i)\\ \\ \dfrac{i}{zw}=-\dfrac{-1+4i}{17}\cdot (1+2i)\\ \\ \dfrac{i}{zw}=-\dfrac{-1+4i-2i-8}{17}\\ \\ \dfrac{i}{zw}=-\dfrac{-9+2i}{17}\\ \\ \dfrac{i}{zw}=\dfrac{9}{17}-\dfrac{2}{17}i\\ \\ \boxed{\mbox{Im}\left(\dfrac{i}{zw}\right)=-\dfrac{2}{17}}$

$\textcircled{3}\\ \\ z+w=(-4+i)+\left(\dfrac{1}{5}-\dfrac{2}{5}i\right)\\ \\ z+w=-\dfrac{19}{5}+\dfrac{3}{5}i\\ \\ |z+w|=\sqrt{\left(-\dfrac{19}{5}\right)^2+\left(\dfrac{3}{5}\right)^2}\\ \\ |z+w|=\sqrt{\dfrac{361}{25}+\dfrac{9}{25}}\\ \\\\ |z+w|=\sqrt{\dfrac{370}{25}}\\ \\ \\ \boxed{|z+w|=\dfrac{\sqrt{370}}{5}}$