Question

Match the products of complex numbers with their standard forms.

(5-2i)(4-i) (2-5i)(1-4i) (1-4i)(5-2i) (4+i)(2-5i)

Answer 1

1)
`(5-2i)(4-i)=18-13i`

2)
`(2-5i)(1-4i)=-18-13i`

3)
`(1-4i)(5-2i)=-3-22i`

4)
`(4+i)(2-5i)=13-18i`

Explanation:

Given : Complex numbers

1) (5-2i)(4-i)

2) (2-5i)(1-4i)

3) (1-4i)(5-2i)

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

To find : The products of complex numbers with their standard forms.

Solution :

We simply apply multiplicative property of brackets,

i.e,
`(a+b)(c+d)=ac+bc+bc+bd`

Note -
`i^2=(√(-1))^2=-1`

1) (5-2i)(4-i)

`(5-2i)(4-i)=((5)(4)+(-2i)(4)+(-5)(i)+(-2i)(-i))`

`(5-2i)(4-i)=(20-8i-5i+2i^2)`

`(5-2i)(4-i)=(20-13i-2)`

`(5-2i)(4-i)=18-13i`

2) (2-5i)(1-4i)

`(2-5i)(1-4i)=((2)(1)+(-5i)(1)+(2)(-4i)+(-5i)(-4i))`

`(2-5i)(1-4i)=(2-5i-8i+20i^2)`

`(2-5i)(1-4i)=(2-13i-20)`

`(2-5i)(1-4i)=-18-13i`

3) (1-4i)(5-2i)

`(1-4i)(5-2i)=((5)(1)+(-4i)(5)+(-2i)(1)+(-2i)(-4i))`

`(1-4i)(5-2i)=(5-20i-2i+8i^2)`

`(1-4i)(5-2i)=(5-22i-8)`

`(1-4i)(5-2i)=-3-22i`

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

`(4+i)(2-5i)=((4)(2)+(i)(2)+(4)(-5i)+(i)(-5i))`

`(4+i)(2-5i)=(8+2i-20i-5i^2)`

`(4+i)(2-5i)=(8-18i+5)`

`(4+i)(2-5i)=13-18i`

Therefore, following above are the required results.

Answer 2

( 5 - 2i ) ( 4 - i ) 18 - 13i

( 2 - 5i ) ( 1 - 4i ) -18 - 13i

( 1 - 4i ) ( 5 - 2i ) -3 - 22i

( 4 + i ) ( 2 - 5i ) 13 - 18i