Resolver ((4-3i)^-1+(-2+i)^-1)

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

Calcular

$$-\frac{6}{25}-\frac{2}{25}i=-0.24-0.08i$$

Pasos da solución

Calcula 4-3i á potencia de -1 e obtén \frac{4}{25}+\frac{3}{25}i.

$$\frac{4}{25}+\frac{3}{25}i+\left(-2+i\right)^{-1}$$

Calcula -2+i á potencia de -1 e obtén -\frac{2}{5}-\frac{1}{5}i.

$$\frac{4}{25}+\frac{3}{25}i+\left(-\frac{2}{5}-\frac{1}{5}i\right)$$

Suma \frac{4}{25}+\frac{3}{25}i e -\frac{2}{5}-\frac{1}{5}i para obter -\frac{6}{25}-\frac{2}{25}i.

$$-\frac{6}{25}-\frac{2}{25}i$$

Parte real

$$-\frac{6}{25} = -0.24$$

Pasos da solución

Calcula 4-3i á potencia de -1 e obtén \frac{4}{25}+\frac{3}{25}i.

$$Re(\frac{4}{25}+\frac{3}{25}i+\left(-2+i\right)^{-1})$$

Calcula -2+i á potencia de -1 e obtén -\frac{2}{5}-\frac{1}{5}i.

$$Re(\frac{4}{25}+\frac{3}{25}i+\left(-\frac{2}{5}-\frac{1}{5}i\right))$$

Suma \frac{4}{25}+\frac{3}{25}i e -\frac{2}{5}-\frac{1}{5}i para obter -\frac{6}{25}-\frac{2}{25}i.

$$Re(-\frac{6}{25}-\frac{2}{25}i)$$

A parte real de -\frac{6}{25}-\frac{2}{25}i é -\frac{6}{25}.

$$-\frac{6}{25}$$