Respuesta :

¡Hola!

[tex] \mathbb{z} = (3 + 4i)(1 + i) \\ \\ 3 (1 + i) = 3 + 3i \\ 4i(1 + i) = 4i + 4 {i}^{2} \\ {i}^{2} = - 1 \\ \\ \mathbb{z} = 3 + 3i + 4i + 4( - 1) \\ \\ \mathbb{z} = 3 - 4+ 7i \\ \\ \mathbb{z} = - 1 + 7i[/tex]

Calculamos su módulo...

[tex] | \mathbb{z}| = \sqrt{ {a}^{2} + {b}^{2} } \\ \\ | \mathbb{z}| = \sqrt{( - 1 {)}^{2} + (7i {)}^{2} } \\ \\ | \mathbb{z}| = \sqrt{1 + 49 {i}^{2} } \\ \\ | \mathbb{z} | = \sqrt{1 + 49( - 1)} \\ \\ | \mathbb{z}| = \sqrt{1 - 49} \\ \\ | \mathbb{z}| = \sqrt{ - 48} = \sqrt{48} \: \sqrt{ - 1} \\ \\ | \mathbb{z}| = i \sqrt{ {2}^{4} \times 3} = i \sqrt{ {2}^{2} \times {2}^{2} \times 3 } \\ \\ | \mathbb{z}| = 2 \times 2 \sqrt{3} i \\ \\ \boxed{ \boxed{| \mathbb{z}| = 4 \sqrt{3} i}}[/tex]

Espero que sirva y saludos.

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