Help With Imaginary Numbers?

Question

Q- The expression (2+i) / (3+i) is equivelant to: (Please try to show work, or explain)

A- (6+5i) / 8
B-(6+i) /8
C- (7-5i) /10
D- (7+i) /10
10 pts for best answer!!! I reealllly need help!

Answer 1

To rationalise a complex denominator, multiply top and bottom by the complex conjugate (same real part, negated imaginary part):
(2 + i) / (3 + i)
= (2 + i) (3 - i) / [(3 + i) (3 - i)]
= (6 - 2i + 3i - i^2) / (9 - i^2)
= (6 + i + 1) / (9 + 1)
= (7 + i) / 10.

Answer 2

You need to multiply the expression by the complex conjugate of the denominator. In this case you multiply the numerator and the denominator by 3 - i. (This is equivalent to multiplying by 1 is it is a-okay!)

The numerator will become: (2 + i)(3 - i) = 6 + i +1 = 7 + i

The denominator will become (3 + i)(3 - i) = 9 + 1 = 10

The expression become (7 + i)/10.

The answer is D.

Answer 3

The answer is D because
(2+i)/(3+i) = ((2+i)*(3-i))/((3+i)*(3-i)), in other words multiply by (3-i)/(3-i). Now you will have when you multiply out the denominator and numerator, (6-2i+3i-i^2)/(9+3i-3i-i^2) and i^2 = -1. so you have (7+i)/10

Answer 4

[(2+i) / (3+i)] * [(3-i) / (3-i)]
(6 + i - i^2) / (9 + i^2)
i^2 = -1, so
(7 + i) / (10)
so D

Answer 5

(2+i)(3-i)/(3+i)(3-i)
=6+1+i/9+1)
=(7/10)+(1/10)i

Answer 6

(7+i)/10

Answer 7

I just can't Imagine.