How do you multiply #(-8i)^2#?

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Answer 1 · 2016-08-01T11:03:00

#-64#

Let's do this in increasingly complex bits:

There are 3 elements to what is being squared: #8, -1, and i#.

If we were to only have #8^2#, we'd know we'd have #8*8=64#

If we were to have #(-8)^2#, we'd know we'd have #8*8*-1*-1=64#

But we have #(-8i)^2#, which means we have #8*8*-1*-1*i*i#.

We already know that the first four terms product to 64. So what to do with the #i#?

#i=sqrt(-1)#, which means that #i^2=i*i=sqrt(-1)*sqrt(-1)=-1#.

And so now we can work out #(-8i)^2 = -64#.