Question

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

Answer 1

Use the formula
`(a+b)(a-b)=a^2-b^2`.


`(4+3i)(4-3i)=4^2-(3i)^2=16-9i^2 \\ \\ i^2=-1 \\ \\ 16-9i^2=16-9 * (-1)=16+9=25 \\ \\ \boxed{(4+3i)(4-3i)=25}`

Answer 2

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