+0  
 
0
27
1
avatar

Hey, was wondering if anyone could help me with this

 

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

 
 Sep 21, 2022
 #1
avatar
+1

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!

 
 Sep 21, 2022

12 Online Users