Express the product (3-i) (4-i)?

Question

in the form of a + bi

second question

Simplify in a + bi form

(4 -i)^2


As you could see I am not understanding the a + bi form

Answer 1

It works just like FOIL. Except that you need to remember that i² becomes -1.

(3 - i)(4 - i) = 12 - 3i - 4i + i² = 12 - 7i + (-1) = 11 - 7i

(4 - i)² = 16 - 8i + i² = 16 - 8i - 1 = 15 - 8i

Answer 2

You should consider a+b*i as an binomial expression remembering tht i^2=-1
(3-i)*(4-i) = 12 -3i -4i +i^2 =11-7i
(4-i)^2= 16 -8i +i^2 = 15-8i

Answer 3

a is a constant and b is the coefficient for the i term. Do the usual foiling of (3 - i)(4 - i), then change the iexp2 term to a whole number (remember that iexp2 = -1).

EX: (7 - i)exp2
(7 - i)(7 - i)= 49 -7i -7i + iexp2
iexp2 = -1 and -7i - 7i = -14i
49 -1 -14i = 48 -14i (48=a, -14=b)

Answer 4

first one:
(3-i)(4-i) = 12 - 7i +i^2
but we know i^2 = -1 by definition of i = the sqrt of -1
so it equals 12 - 7i + -1
or 11 - 7i

second:
(4-i)(4-i) = 16 - 8i + i^2
=16 - 8i + -1
=15 - 8i

Answer 5

(3-i)(4-i)
= 12-7i+i^2
------> i=sqrt(-1)
12-7i+i^2
=12 - 1 -7i
=11-7i


(4-i)^2
=16 - 8i + i^2
=16 - 8i -1
=15-8i