Hey, was wondering if anyone could help me with this
Simplify $(1-3i)(1-i)(1+i)(1+3i)$.
Recall that: \((a-b)(a+b)=a^2-b^2\) for any a and b. (Difference between two squares).
So, rearrange the given question:
\((1-3i)(1+3i)(1-i)(1+i)\)
So, using the above formula:
\((1^2-(3i)^2)(1^2-(i)^2)=(1-9i^2)(1-i^2) \\ \text{But, } i^2=-1 \hspace{0.4cm} \text{ so:} \\ (1+9)(1+1)=10(2)=20\)
I hope this helps!
