Prove the indicated operations.
(3-i)(4+3i)(5-2i)

I hope if one of these help.

Step-by-step explanation:

(4 - 3i) + (2 + i )(5 - 2i)

=16 - 2i

If you simplify your answer will be: 16−2i

1) (3+2i)-(1-6i)

2) 4i+3-6+i-1

3) 5i(3-2i)

4) (5+3i)(2-3i)

5) (2+i)(2-i)

1) (3 + 2i) - (1 - 6i)

 3 + 2i - 1 + 6i

 2 + 8i

2) 4i + 3 - 6 + i - 1

  5i - 3 - 1

  5i - 4

3) 5i ( 3 - 2i )

  15i - 10i^2

  15i - 10 ( -1 )

  15i + 10

4) ( 5 + 3i ) ( 2 - 3i )

  10 - 15i + 6i - 9i^2

  10 - 9i - 9i^2

  10 - 9i - 9 ( -1 )

  10 - 9i + 9

 - 9i + 19

5) ( 2 + i ) ( 2 - i )

   4  - 2i + 2i - i^2

   4 - i^2

   4 - (-1)

   5

1) 2 + 8i

2) -4 + 5i

3) 10 + 15i

4) 19 - 9i

5) 5

The operations with imaginary numbers are equal to those of the real numbers considering i as a variable, but with a property that allows to simplify the expression more, that is that i squared equals -1. Taking this property into account, we solve the expressions like this:

1) (3 + 2i) - (1-6i) = 3 + 2i - 1 + 6i = 2 + 8i

2) 4i + 3-6 + i-1 = 5i - 4 = -4 + 5i

3) 5i (3-2i) = 5i*3 - 5i*2i = 15i - 10i^2 = 15i - 10(-1) = 15i + 10 = 10 + 15i

4) (5 + 3i) (2-3i) = 5*2 - 5 3i + 3i*2 - 3*3i^2 = 10 - 15i + 6i - 9i^2 = 10 - 9i - 9(-1) = 10 - 9i + 9 = 19 - 9i

5) (2 + i) (2-i) = 2*2 - 2*i + 2*i - i^2 = 4 - 2i + 2i - (-1) = 4 + 1 = 5

I hope one of these help:)

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Step-by-step explanation:

This is formula fot completing the square:

(b/2)^(2)

(-20/2)^(2)

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