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# Moooooth question

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Hey, was wondering if anyone could help me with this

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

Sep 21, 2022

#1
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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