( 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:
1) [tex](5-2i)(4-i)=18-13i[/tex]
2) [tex](2-5i)(1-4i)=-18-13i[/tex]
3) [tex](1-4i)(5-2i)=-3-22i[/tex]
4) [tex](4+i)(2-5i)=13-18i[/tex]
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, [tex](a+b)(c+d)=ac+bc+bc+bd[/tex]
Note - [tex]i^2=(\sqrt{-1})^2=-1[/tex]
1) (5-2i)(4-i)
[tex](5-2i)(4-i)=((5)(4)+(-2i)(4)+(-5)(i)+(-2i)(-i))[/tex]
[tex](5-2i)(4-i)=(20-8i-5i+2i^2)[/tex]
[tex](5-2i)(4-i)=(20-13i-2)[/tex]
[tex](5-2i)(4-i)=18-13i[/tex]
2) (2-5i)(1-4i)
[tex](2-5i)(1-4i)=((2)(1)+(-5i)(1)+(2)(-4i)+(-5i)(-4i))[/tex]
[tex](2-5i)(1-4i)=(2-5i-8i+20i^2)[/tex]
[tex](2-5i)(1-4i)=(2-13i-20)[/tex]
[tex](2-5i)(1-4i)=-18-13i[/tex]
3) (1-4i)(5-2i)
[tex](1-4i)(5-2i)=((5)(1)+(-4i)(5)+(-2i)(1)+(-2i)(-4i))[/tex]
[tex](1-4i)(5-2i)=(5-20i-2i+8i^2)[/tex]
[tex](1-4i)(5-2i)=(5-22i-8)[/tex]
[tex](1-4i)(5-2i)=-3-22i[/tex]
4) (4+i)(2-5i)
[tex](4+i)(2-5i)=((4)(2)+(i)(2)+(4)(-5i)+(i)(-5i))[/tex]
[tex](4+i)(2-5i)=(8+2i-20i-5i^2)[/tex]
[tex](4+i)(2-5i)=(8-18i+5)[/tex]
[tex](4+i)(2-5i)=13-18i[/tex]
Therefore, following above are the required results.
