Question

1) Calcule:

a)  $\frac{2+i}{3-2i}$
 
b) $\frac{2+3i}{2-i}- \frac{1-i}{2+i}$

c) $\frac{2i}{4-2i}$

d) $\frac{1-2i}{3+i}- \frac{i}{1-4i}$

Answer 1

a) $\dfrac{2+i}{3-2i}=\dfrac{2+i}{3-2i}\cdot\dfrac{3+2i}{3+2i}=\dfrac{(2+i)(3+2i)}{(3-2i)(3+2i)}$

$\dfrac{2+i}{3-2i}=\dfrac{6+4i+3i+2i^2}{3^2-(2i)^2}=\dfrac{6+7i-2}{9-4i^2}=\dfrac{4+7i}{9+4}=\dfrac{4+7i}{13}$

b) $\dfrac{2+3i}{2-i}-\dfrac{1-i}{2+i}$.

Vamos por partes:

$\dfrac{2+3i}{2-i}=\dfrac{(2+3i)(2+i)}{(2-i)(2+i)}=\dfrac{4+2i+6i+3i^2}{2^2-i^2}$

$\dfrac{2+3i}{2-i}=\dfrac{4+8i-3}{4+1}=\dfrac{1+8i}{5}$


$\dfrac{1-i}{2+i}=\dfrac{(1-i)(2-i)}{(2+i)(2-i)}=\dfrac{2-i-2i+i^2}{2^2-i^2}=\dfrac{2-3i-1}{4+1}$

$\dfrac{1-i}{2+i}=\dfrac{1-3i}{5}$

$\dfrac{2+3i}{2-i}-\dfrac{1-i}{2+i}=\dfrac{1+8i}{5}-\dfrac{1-3i}{5}=\dfrac{1+8i-1+3i}{5}=\dfrac{11i}{5}$


c) $\dfrac{2i}{4-2i}=\dfrac{2i(4+2i)}{(4-2i)(4+2i)}=\dfrac{8i+4i^2}{4^2-(2i)^2}=\dfrac{8i-4}{16+4i^2}$

$\dfrac{2i}{4-2i}=\dfrac{8i-4}{16-4}=\dfrac{8i-4}{12}=\dfrac{2i-1}{3}$


d) $\dfrac{1-2i}{3+i}-\dfrac{i}{1-4i}$

Vamos por partes:

$\dfrac{1-2i}{3+i}=\dfrac{(1-2i)(3-i)}{(3+i)(3-i)}=\dfrac{3-i-6i+2i^2}{3^2-i^2}=\dfrac{3-7i-2}{9+1}$

$\dfrac{1-2i}{3+i}=\dfrac{1-7i}{10}$

$\dfrac{i}{1-4i}=\dfrac{i(1+4i)}{(1-4i)(1+4i)}=\dfrac{i+4i^2}{1^2-(4i)^2}=\dfrac{i-4}{1-16i^2}$

$\dfrac{i}{1-4i}=\dfrac{i-4}{1+16}=\dfrac{i-4}{17}$.

$\dfrac{1-2i}{3+i}-\dfrac{i}{1-4i}=\dfrac{1-7i}{10}-\dfrac{i-4}{17}=\dfrac{17(1-7i)-10(i-4)}{170}=\dfrac{17-119i-10i+40}{170}$

$\dfrac{1-2i}{3+i}-\dfrac{i}{1-4i}=\dfrac{57-129i}{170}$.