How do i Simplify (4+3i)(4 + 3i)?

(4+3i)(4 + 3i) = 16+12i+12i+9i^2 where i = sqrt(-1)

= 16+24i+9(-1)

= 7+24i

Multiply each number in the first set with the numbers in the second set. So = 4*4 + 4*3i + 3i*4 + 3i*3i and then simplify the result as you normally would. Or if you just want to write the equation in shorter form it is: (4 + 3i)^2

You'd simplify this the way you would any other binomial pair. Remember that (3i)²= (3²)(i²) = 9(-1) = -9

(4 + 3i)² = 16 + 12i + 12i - 9

= 7 + 24i

(4+3i)(4 + 3i) = (4 + 3i)^2 = 16 +24i + 9i^2 (using this rule: (a+b)^2 = a^2 + 2ab + b^2))

now i^2 = -1

so

16 + 24i -9

7 + 24i

Hope this helps =)

✐Explanation✐

Make i as an imaginary number.

i = i

i² = -1

i³ = -i

i^4 = 1

Multiply each term by term and combine like terms. Then, simplify.

16 + 12i + 12i + 9i²

16 + 24i + 9i²

16 + 24i - 9

7 + 24i

You use the FOIL method.

First, Outside, Inside, Last.

F- 4x4= 16

O- 4x3i= 12i

I- 3ix4= 12i

L- 3ix3i= 9i(squared2)

then you combine like terms:

12i + 12i= 34i.

So the answer would be

9i(squared2)+34i+16

:)

16 - 9 i ²

16 + 9

25

(4 + 3i)(4 + 3i) = 4(4 + 3i) + 3i(4 + 3i)

...................... = 16 + 12i + 12i + 9i²

...................... = 16 + 12i + 12i + 9(–1)

...................... = 16 + 12i + 12i – 9

...................... = 7 + 24i