Resolver (4-i)(6+2i)-(7-i)(4+3i)

$$( 4 - i ) ( 6 + 2 i ) - ( 7 - i ) ( 4 + 3 i )$$

Calcular

$$-5-15i$$

Pasos da solución

Multiplica os números complexos 4-i e 6+2i igual que se multiplican os binomios.

$$4\times 6+4\times \left(2i\right)-i\times 6-2i^{2}-\left(7-i\right)\left(4+3i\right)$$

Por definición, i^{2} é -1.

$$4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right)-\left(7-i\right)\left(4+3i\right)$$

Fai as multiplicacións en 4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right).

$$24+8i-6i+2-\left(7-i\right)\left(4+3i\right)$$

Combina as partes reais e imaxinarias en 24+8i-6i+2.

$$24+2+\left(8-6\right)i-\left(7-i\right)\left(4+3i\right)$$

Fai as sumas en 24+2+\left(8-6\right)i.

$$26+2i-\left(7-i\right)\left(4+3i\right)$$

Multiplica os números complexos 7-i e 4+3i igual que se multiplican os binomios.

$$26+2i-\left(7\times 4+7\times \left(3i\right)-i\times 4-3i^{2}\right)$$

Por definición, i^{2} é -1.

$$26+2i-\left(7\times 4+7\times \left(3i\right)-i\times 4-3\left(-1\right)\right)$$

Fai as multiplicacións en 7\times 4+7\times \left(3i\right)-i\times 4-3\left(-1\right).

$$26+2i-\left(28+21i-4i+3\right)$$

Combina as partes reais e imaxinarias en 28+21i-4i+3.

$$26+2i-\left(28+3+\left(21-4\right)i\right)$$

Fai as sumas en 28+3+\left(21-4\right)i.

$$26+2i-\left(31+17i\right)$$

Para restar 31+17i de 26+2i, resta as súas partes reais e imaxinarias correspondentes.

$$26-31+\left(2-17\right)i$$

Resta 31 de 26. Resta 17 de 2.

$$-5-15i$$

Parte real

$$-5$$

Pasos da solución

Multiplica os números complexos 4-i e 6+2i igual que se multiplican os binomios.

$$Re(4\times 6+4\times \left(2i\right)-i\times 6-2i^{2}-\left(7-i\right)\left(4+3i\right))$$

Por definición, i^{2} é -1.

$$Re(4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right)-\left(7-i\right)\left(4+3i\right))$$

Fai as multiplicacións en 4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right).

$$Re(24+8i-6i+2-\left(7-i\right)\left(4+3i\right))$$

Combina as partes reais e imaxinarias en 24+8i-6i+2.

$$Re(24+2+\left(8-6\right)i-\left(7-i\right)\left(4+3i\right))$$

Fai as sumas en 24+2+\left(8-6\right)i.

$$Re(26+2i-\left(7-i\right)\left(4+3i\right))$$

Multiplica os números complexos 7-i e 4+3i igual que se multiplican os binomios.

$$Re(26+2i-\left(7\times 4+7\times \left(3i\right)-i\times 4-3i^{2}\right))$$

Por definición, i^{2} é -1.

$$Re(26+2i-\left(7\times 4+7\times \left(3i\right)-i\times 4-3\left(-1\right)\right))$$

Fai as multiplicacións en 7\times 4+7\times \left(3i\right)-i\times 4-3\left(-1\right).

$$Re(26+2i-\left(28+21i-4i+3\right))$$

Combina as partes reais e imaxinarias en 28+21i-4i+3.

$$Re(26+2i-\left(28+3+\left(21-4\right)i\right))$$

Fai as sumas en 28+3+\left(21-4\right)i.

$$Re(26+2i-\left(31+17i\right))$$

Para restar 31+17i de 26+2i, resta as súas partes reais e imaxinarias correspondentes.

$$Re(26-31+\left(2-17\right)i)$$

Resta 31 de 26. Resta 17 de 2.

$$Re(-5-15i)$$

A parte real de -5-15i é -5.

$$-5$$