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what is the product of 4+3i and 4-3i

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Naǫid
Use the formula [tex](a+b)(a-b)=a^2-b^2[/tex].

[tex](4+3i)(4-3i)=4^2-(3i)^2=16-9i^2 \\ \\ i^2=-1 \\ \\ 16-9i^2=16-9 \times (-1)=16+9=25 \\ \\ \boxed{(4+3i)(4-3i)=25}[/tex]
Apologiabiologyid
so we know that i=the square root of -1 so
we multiply and notice that is is in the format of the differnece of two perfect squares which is (x)^2-(y)^2=(x-y)(x+y) so
we notice that 4=x and 3i=y so

(4-3i)(4+3i)=(4)^2-(3i)^2

(3 times the square root of -3)^2=(9 times -1)=-9
4^2 is 16
16-(-9)=subtracting a negative=adding a positive so =16+9=25
the answer is 25

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