Resolver (1-2i)*(-5+4i)+(-2-2i)

$$( 1 - 2 i ) \cdot ( - 5 + 4 i ) + ( - 2 - 2 i )$$

Calcular

$$1+12i$$

Pasos da solución

Multiplica os números complexos 1-2i e -5+4i igual que se multiplican os binomios.

$$1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4i^{2}+\left(-2-2i\right)$$

Por definición, i^{2} é -1.

$$1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4\left(-1\right)+\left(-2-2i\right)$$

Fai as multiplicacións en 1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4\left(-1\right).

$$-5+4i+10i+8+\left(-2-2i\right)$$

Combina as partes reais e imaxinarias en -5+4i+10i+8.

$$-5+8+\left(4+10\right)i+\left(-2-2i\right)$$

Fai as sumas en -5+8+\left(4+10\right)i.

$$3+14i+\left(-2-2i\right)$$

Combina as partes reais e imaxinarias.

$$3-2+\left(14-2\right)i$$

Fai as sumas.

$$1+12i$$

Parte real

$$1$$

Pasos da solución

Multiplica os números complexos 1-2i e -5+4i igual que se multiplican os binomios.

$$Re(1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4i^{2}+\left(-2-2i\right))$$

Por definición, i^{2} é -1.

$$Re(1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4\left(-1\right)+\left(-2-2i\right))$$

Fai as multiplicacións en 1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4\left(-1\right).

$$Re(-5+4i+10i+8+\left(-2-2i\right))$$

Combina as partes reais e imaxinarias en -5+4i+10i+8.

$$Re(-5+8+\left(4+10\right)i+\left(-2-2i\right))$$

Fai as sumas en -5+8+\left(4+10\right)i.

$$Re(3+14i+\left(-2-2i\right))$$

Combina as partes reais e imaxinarias en 3+14i+\left(-2-2i\right).

$$Re(3-2+\left(14-2\right)i)$$

Fai as sumas en 3-2+\left(14-2\right)i.

$$Re(1+12i)$$

A parte real de 1+12i é 1.

$$1$$