Question

Match the products of complex numbers with their standard forms.

Answer 1

( 5 - 2i ) ( 4 - i )       18 - 13i

( 2 - 5i ) ( 1 - 4i )     -18 - 13i

( 1 - 4i ) ( 5 - 2i )      -3 - 22i

( 4 + i ) ( 2 - 5i )     13 - 18i

Answer 2

Answer:

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

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

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

4) $(4+i)(2-5i)=13-18i$    

Step-by-step 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 - $i^2=(\sqrt{-1})^2=-1$

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.