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MrMuchimi
3 years ago
14

Compute the special products. (-3+5i)2

Mathematics
1 answer:
sveticcg [70]3 years ago
4 0

we are given

(-3+5i)^2

we can expand it

(-3+5i)^2=(-3+5i)(-3+5i)

now, we can FOIL it

(-3+5i)^2=-3\times -3-3\times 5i+5i\times -3+5i\times 5i

now, we can simplify it

(-3+5i)^2=9-15i-15i+25i^2

we know that

i^2=-1

so, we can plug it

(-3+5i)^2=9-30i+25(-1)

(-3+5i)^2=9-30i-25

(-3+5i)^2=-16-30i..............Answer


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