Question

Simplify $(1-3i)(1-i)(1+i)(1+3i)$

Answer 1

Answer:

$\huge\boxed{(1-3i)(1-i)(1+i)(1+3i)=20}$

Step-by-step explanation:

$(1-3i)(1-i)(1+i)(1+3i)\\\\\text{use the commutative property}\\\\=(1-3i)(1+3i)(1-i)(1+i)\\\\\text{use the associative property}\\\\=\bigg[(1-3i)(1+3i)\bigg]\bigg[(1-i)(1+i)\bigg]\\\\\text{use}\ (a-b)(a+b)=a^2-b^2\\\\=\bigg[1^2-(3i)^2\bigg]\bigg[1^2-i^2\bigg]\\\\=\bigg(1-9i^2\bigg)\bigg(1-i^2\bigg)\\\\\text{use}\ i=\sqrt{-1}\to i^2=-1\\\\=\bigg(1-9(-1)\bigg)\bigg(1-(-1)\bigg)\\\\=\bigg(1+9\bigg)\bigg(1+1\bigg)\\\\=(10)(2)\\\\=20$

Answer 2

$(1-3i)(1-i)(1+i)(1+3i)=\\(1^2-(3i)^2)(1^2-i^2)=\\(1+9)(1+1)=\\10\cdot2=20$