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

Answer 1

I found: #60-32i#

We can rite it as: #(8-2i)(8-2i)=(8*8)-(2*8i)-(2*8i)+(2*2*i*i)=64-16i-16i+4i^2=# but #i^2=(sqrt(-1))^2=-1# so: #=64-32i-4=60-32i#
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Answer 2

To simplify ( (8-2i)^2 ), you can use the FOIL method, which stands for First, Outer, Inner, Last.

First, multiply the first terms of each binomial: ( 8 \times 8 = 64 ).

Outer, multiply the outer terms of each binomial: ( 8 \times (-2i) = -16i ).

Inner, multiply the inner terms of each binomial: ( (-2i) \times 8 = -16i ).

Last, multiply the last terms of each binomial: ( (-2i) \times (-2i) = 4i^2 ).

Now, combine the results: ( 64 - 16i - 16i + 4i^2 ).

Remember that ( i^2 = -1 ), so ( 4i^2 = 4(-1) = -4 ).

Combine like terms: ( 64 - 32i - 4 ).

Finally, simplify: ( 60 - 32i ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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