The product of (3 + 2i) and a complex number is (17 + 7i).

1 answer:

8 0

A+bi is a complex number

(3+2i)(a+bi)=17+7i
remember
i²=-1
so
expand and solve
a is the ral part
bi is the imaginary part
ok so
if Ac+Be=dc+fe where c=c and e=e then A=d and B=f

(3+2i)(a+bi)=17+7i
expand/distribute/FOIL
3a+2ai+3bi+2bi²=17+7i
3a+2bi+2ai+2bi=17+7i
3a-2b+2ai+2bi=17+7i
real parts are 3a-2b
imaginary is 2ai+2bi
so

3a-2b=17 and
2ai+2bi=7i

we need to solve
2nd equation, divide both sides by i
2a+2b=7
multiply by -1 and add to other equation

3a+2b=17
<u>-2a-2b=-7 +</u>
1a+0b=10
a=10

subsitue
2a+2b=7
2(10)+2b=7
20+2b=7
2b=-13
b=-13/2

the complex number is
10-(13/2)i or
10-6.5i

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