Question

(NuMeros complexos) Calcule; 3+5i/4-2i




E se possível
2+3i/3+i

Answer 1

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

Multiplicando o numerador e o denominador da fração por 4 + 2i (conjugado de 4 - 2i):

$\dfrac{3+5i}{4-2i}=\dfrac{(3+5i)(4+2i)}{(4-2i)(4+2i)}\\\\\\\dfrac{3+5i}{4-2i}=\dfrac{12+6i+20i+10i^{2}}{4^{2}-(2i)^{2}}\\\\\\\dfrac{3+5i}{4-2i}=\dfrac{12+26i+10(-1)}{16-2^{2}i^{2}}\\\\\\\dfrac{3+5i}{4-2i}=\dfrac{12+26i-10}{16-4(-1)}\\\\\\\dfrac{3+5i}{4-2i}=\dfrac{2+26i}{16+4}\\\\\\\dfrac{3+5i}{4-2i}=\dfrac{2+26i}{20}\\\\\\\dfrac{3+5i}{4-2i}=\dfrac{1+13i}{10}\\\\\\\boxed{\boxed{\dfrac{3+5i}{4-2i}=\dfrac{1}{10}+\dfrac{13}{10}i}}$
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$\dfrac{2+3i}{3+i}=\dfrac{(2+3i)(3-i)}{(3+i)(3-i)}\\\\\\\dfrac{2+3i}{3+i}=\dfrac{6-2i+9i-3i^{2}}{3^{2}-i^{2}}\\\\\\\dfrac{2+3i}{3+i}=\dfrac{6+7i-3(-1)}{9-(-1)}\\\\\\\dfrac{2+3i}{3+i}=\dfrac{6+7i+3}{9+1}\\\\\\\dfrac{2+3i}{3+i}=\dfrac{9+7i}{10}\\\\\\\boxed{\boxed{\dfrac{2+3i}{3+i}=\dfrac{9}{10}+\dfrac{7}{10}i}}$